677 lines
19 KiB
Markdown
677 lines
19 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 404.左叶子之和
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[力扣题目链接](https://leetcode.cn/problems/sum-of-left-leaves/)
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计算给定二叉树的所有左叶子之和。
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示例：
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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：：[二叉树的题目中，总有一些规则让你找不到北 | LeetCode：404.左叶子之和](https://www.bilibili.com/video/BV1GY4y1K7z8)，相信结合视频在看本篇题解，更有助于大家对本题的理解**。
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## 思路
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**首先要注意是判断左叶子，不是二叉树左侧节点，所以不要上来想着层序遍历。**
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因为题目中其实没有说清楚左叶子究竟是什么节点，那么我来给出左叶子的明确定义：**节点A的左孩子不为空，且左孩子的左右孩子都为空（说明是叶子节点），那么A节点的左孩子为左叶子节点**
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大家思考一下如下图中二叉树，左叶子之和究竟是多少？
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**其实是0，因为这棵树根本没有左叶子！**
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但看这个图的左叶子之和是多少？
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相信通过这两个图，大家对最左叶子的定义有明确理解了。
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那么**判断当前节点是不是左叶子是无法判断的，必须要通过节点的父节点来判断其左孩子是不是左叶子。**
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如果该节点的左节点不为空，该节点的左节点的左节点为空，该节点的左节点的右节点为空，则找到了一个左叶子，判断代码如下：
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```CPP
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if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
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左叶子节点处理逻辑
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}
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```
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### 递归法
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递归的遍历顺序为后序遍历（左右中），是因为要通过递归函数的返回值来累加求取左叶子数值之和。
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递归三部曲：
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1. 确定递归函数的参数和返回值
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判断一个树的左叶子节点之和，那么一定要传入树的根节点，递归函数的返回值为数值之和，所以为int
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使用题目中给出的函数就可以了。
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2. 确定终止条件
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如果遍历到空节点，那么左叶子值一定是0
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```CPP
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if (root == NULL) return 0;
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```
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注意，只有当前遍历的节点是父节点，才能判断其子节点是不是左叶子。 所以如果当前遍历的节点是叶子节点，那其左叶子也必定是0，那么终止条件为：
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```CPP
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if (root == NULL) return 0;
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if (root->left == NULL && root->right== NULL) return 0; //其实这个也可以不写，如果不写不影响结果，但就会让递归多进行了一层。
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```
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3. 确定单层递归的逻辑
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当遇到左叶子节点的时候，记录数值，然后通过递归求取左子树左叶子之和，和 右子树左叶子之和，相加便是整个树的左叶子之和。
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代码如下：
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```CPP
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int leftValue = sumOfLeftLeaves(root->left); // 左
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if (root->left && !root->left->left && !root->left->right) {
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leftValue = root->left->val;
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}
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int rightValue = sumOfLeftLeaves(root->right); // 右
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int sum = leftValue + rightValue; // 中
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return sum;
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```
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整体递归代码如下：
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```CPP
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class Solution {
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public:
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int sumOfLeftLeaves(TreeNode* root) {
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if (root == NULL) return 0;
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if (root->left == NULL && root->right== NULL) return 0;
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int leftValue = sumOfLeftLeaves(root->left); // 左
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if (root->left && !root->left->left && !root->left->right) { // 左子树就是一个左叶子的情况
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leftValue = root->left->val;
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}
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int rightValue = sumOfLeftLeaves(root->right); // 右
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int sum = leftValue + rightValue; // 中
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return sum;
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}
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};
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```
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以上代码精简之后如下：
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```CPP
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class Solution {
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public:
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int sumOfLeftLeaves(TreeNode* root) {
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if (root == NULL) return 0;
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int leftValue = 0;
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if (root->left != NULL && root->left->left == NULL && root->left->right == NULL) {
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leftValue = root->left->val;
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}
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return leftValue + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
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}
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};
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```
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精简之后的代码其实看不出来用的是什么遍历方式了，对于算法初学者以上根据第一个版本来学习。
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### 迭代法
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本题迭代法使用前中后序都是可以的，只要把左叶子节点统计出来，就可以了，那么参考文章 [二叉树：听说递归能做的，栈也能做！](https://programmercarl.com/二叉树的迭代遍历.html)和[二叉树：迭代法统一写法](https://programmercarl.com/二叉树的统一迭代法.html)中的写法，可以写出一个前序遍历的迭代法。
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判断条件都是一样的，代码如下：
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```CPP
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class Solution {
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public:
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int sumOfLeftLeaves(TreeNode* root) {
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stack<TreeNode*> st;
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if (root == NULL) return 0;
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st.push(root);
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int result = 0;
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while (!st.empty()) {
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TreeNode* node = st.top();
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st.pop();
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if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
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result += node->left->val;
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}
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if (node->right) st.push(node->right);
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if (node->left) st.push(node->left);
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}
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return result;
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}
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};
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```
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## 总结
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这道题目要求左叶子之和，其实是比较绕的，因为不能判断本节点是不是左叶子节点。
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此时就要通过节点的父节点来判断其左孩子是不是左叶子了。
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**平时我们解二叉树的题目时，已经习惯了通过节点的左右孩子判断本节点的属性，而本题我们要通过节点的父节点判断本节点的属性。**
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希望通过这道题目，可以扩展大家对二叉树的解题思路。
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## 其他语言版本
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### Java：
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**递归**
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```java
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class Solution {
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public int sumOfLeftLeaves(TreeNode root) {
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if (root == null) return 0;
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int leftValue = sumOfLeftLeaves(root.left); // 左
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int rightValue = sumOfLeftLeaves(root.right); // 右
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int midValue = 0;
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if (root.left != null && root.left.left == null && root.left.right == null) {
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midValue = root.left.val;
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}
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int sum = midValue + leftValue + rightValue; // 中
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return sum;
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}
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}
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```
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**迭代**
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```java
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class Solution {
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public int sumOfLeftLeaves(TreeNode root) {
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if (root == null) return 0;
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Stack<TreeNode> stack = new Stack<> ();
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stack.add(root);
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int result = 0;
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while (!stack.isEmpty()) {
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TreeNode node = stack.pop();
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if (node.left != null && node.left.left == null && node.left.right == null) {
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result += node.left.val;
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}
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if (node.right != null) stack.add(node.right);
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if (node.left != null) stack.add(node.left);
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}
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return result;
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}
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}
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```
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```java
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// 层序遍历迭代法
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class Solution {
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public int sumOfLeftLeaves(TreeNode root) {
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int sum = 0;
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if (root == null) return 0;
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Queue<TreeNode> queue = new LinkedList<>();
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queue.offer(root);
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while (!queue.isEmpty()) {
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int size = queue.size();
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while (size -- > 0) {
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TreeNode node = queue.poll();
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if (node.left != null) { // 左节点不为空
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queue.offer(node.left);
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if (node.left.left == null && node.left.right == null){ // 左叶子节点
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sum += node.left.val;
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}
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}
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if (node.right != null) queue.offer(node.right);
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}
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}
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return sum;
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}
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}
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```
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### Python:
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递归
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def sumOfLeftLeaves(self, root):
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if root is None:
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return 0
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if root.left is None and root.right is None:
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return 0
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leftValue = self.sumOfLeftLeaves(root.left) # 左
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if root.left and not root.left.left and not root.left.right: # 左子树是左叶子的情况
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leftValue = root.left.val
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rightValue = self.sumOfLeftLeaves(root.right) # 右
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sum_val = leftValue + rightValue # 中
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return sum_val
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```
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递归精简版
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def sumOfLeftLeaves(self, root):
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if root is None:
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return 0
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leftValue = 0
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if root.left is not None and root.left.left is None and root.left.right is None:
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leftValue = root.left.val
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return leftValue + self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
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```
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迭代法
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def sumOfLeftLeaves(self, root):
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if root is None:
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return 0
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st = [root]
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result = 0
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while st:
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node = st.pop()
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if node.left and node.left.left is None and node.left.right is None:
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result += node.left.val
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if node.right:
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st.append(node.right)
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if node.left:
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st.append(node.left)
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return result
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```
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### Go:
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**递归法**
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```go
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func sumOfLeftLeaves(root *TreeNode) int {
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if root == nil {
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return 0
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}
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leftValue := sumOfLeftLeaves(root.Left) // 左
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if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {
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leftValue = root.Left.Val // 中
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}
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rightValue := sumOfLeftLeaves(root.Right) // 右
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return leftValue + rightValue
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}
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```
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**迭代法(前序遍历)**
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```go
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func sumOfLeftLeaves(root *TreeNode) int {
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st := make([]*TreeNode, 0)
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if root == nil {
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return 0
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}
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st = append(st, root)
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result := 0
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for len(st) != 0 {
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node := st[len(st)-1]
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st = st[:len(st)-1]
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if node.Left != nil && node.Left.Left == nil && node.Left.Right == nil {
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result += node.Left.Val
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}
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if node.Right != nil {
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st = append(st, node.Right)
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}
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if node.Left != nil {
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st = append(st, node.Left)
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}
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}
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return result
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}
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```
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### JavaScript:
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**递归法**
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```javascript
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var sumOfLeftLeaves = function(root) {
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//采用后序遍历 递归遍历
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// 1. 确定递归函数参数
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const nodesSum = function(node) {
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// 2. 确定终止条件
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if(node === null) {
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return 0;
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}
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let leftValue = nodesSum(node.left);
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let rightValue = nodesSum(node.right);
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// 3. 单层递归逻辑
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let midValue = 0;
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if(node.left && node.left.left === null && node.left.right === null) {
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midValue = node.left.val;
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}
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let sum = midValue + leftValue + rightValue;
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return sum;
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}
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return nodesSum(root);
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};
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```
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**迭代法**
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```javascript
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var sumOfLeftLeaves = function(root) {
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//采用层序遍历
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if(root === null) {
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return null;
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}
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let queue = [];
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let sum = 0;
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queue.push(root);
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while(queue.length) {
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let node = queue.shift();
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if(node.left !== null && node.left.left === null && node.left.right === null) {
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sum+=node.left.val;
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}
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node.left && queue.push(node.left);
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node.right && queue.push(node.right);
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}
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return sum;
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};
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```
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### TypeScript:
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> 递归法
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```typescript
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function sumOfLeftLeaves(root: TreeNode | null): number {
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if (root === null) return 0;
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let midVal: number = 0;
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if (
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root.left !== null &&
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root.left.left === null &&
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root.left.right === null
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) {
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midVal = root.left.val;
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}
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let leftVal: number = sumOfLeftLeaves(root.left);
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let rightVal: number = sumOfLeftLeaves(root.right);
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return midVal + leftVal + rightVal;
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};
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```
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> 迭代法
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```typescript
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function sumOfLeftLeaves(root: TreeNode | null): number {
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let helperStack: TreeNode[] = [];
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let tempNode: TreeNode;
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let sum: number = 0;
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if (root !== null) helperStack.push(root);
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while (helperStack.length > 0) {
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tempNode = helperStack.pop()!;
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if (
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tempNode.left !== null &&
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tempNode.left.left === null &&
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tempNode.left.right === null
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) {
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sum += tempNode.left.val;
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}
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if (tempNode.right !== null) helperStack.push(tempNode.right);
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if (tempNode.left !== null) helperStack.push(tempNode.left);
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}
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return sum;
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};
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```
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### Swift:
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**递归法**
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```swift
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func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
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guard let root = root else {
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return 0
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}
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let leftValue = sumOfLeftLeaves(root.left)
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let rightValue = sumOfLeftLeaves(root.right)
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var midValue: Int = 0
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if root.left != nil && root.left?.left == nil && root.left?.right == nil {
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midValue = root.left!.val
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}
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let sum = midValue + leftValue + rightValue
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return sum
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}
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```
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**迭代法**
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```swift
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func sumOfLeftLeaves(_ root: TreeNode?) -> Int {
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guard let root = root else {
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return 0
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}
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var stack = Array<TreeNode>()
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stack.append(root)
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var sum = 0
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while !stack.isEmpty {
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let lastNode = stack.removeLast()
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if lastNode.left != nil && lastNode.left?.left == nil && lastNode.left?.right == nil {
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sum += lastNode.left!.val
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}
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if let right = lastNode.right {
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stack.append(right)
|
||
}
|
||
if let left = lastNode.left {
|
||
stack.append(left)
|
||
}
|
||
}
|
||
return sum
|
||
}
|
||
```
|
||
|
||
### C:
|
||
递归法：
|
||
```c
|
||
int sumOfLeftLeaves(struct TreeNode* root){
|
||
// 递归结束条件：若当前结点为空，返回0
|
||
if(!root)
|
||
return 0;
|
||
|
||
// 递归取左子树的左结点和和右子树的左结点和
|
||
int leftValue = sumOfLeftLeaves(root->left);
|
||
int rightValue = sumOfLeftLeaves(root->right);
|
||
|
||
// 若当前结点的左结点存在，且其为叶子结点。取它的值
|
||
int midValue = 0;
|
||
if(root->left && (!root->left->left && !root->left->right))
|
||
midValue = root->left->val;
|
||
|
||
return leftValue + rightValue + midValue;
|
||
}
|
||
```
|
||
|
||
迭代法:
|
||
```c
|
||
int sumOfLeftLeaves(struct TreeNode* root){
|
||
struct TreeNode* stack[1000];
|
||
int stackTop = 0;
|
||
|
||
// 若传入root结点不为空，将其入栈
|
||
if(root)
|
||
stack[stackTop++] = root;
|
||
|
||
int sum = 0;
|
||
//若栈不为空，进行循环
|
||
while(stackTop) {
|
||
// 出栈栈顶元素
|
||
struct TreeNode *topNode = stack[--stackTop];
|
||
// 若栈顶元素的左孩子为左叶子结点，将其值加入sum中
|
||
if(topNode->left && (!topNode->left->left && !topNode->left->right))
|
||
sum += topNode->left->val;
|
||
|
||
// 若当前栈顶结点有左右孩子。将他们加入栈中进行遍历
|
||
if(topNode->right)
|
||
stack[stackTop++] = topNode->right;
|
||
if(topNode->left)
|
||
stack[stackTop++] = topNode->left;
|
||
}
|
||
return sum;
|
||
}
|
||
```
|
||
|
||
### Scala:
|
||
|
||
**递归:**
|
||
```scala
|
||
object Solution {
|
||
def sumOfLeftLeaves(root: TreeNode): Int = {
|
||
if(root == null) return 0
|
||
var midValue = 0
|
||
if(root.left != null && root.left.left == null && root.left.right == null){
|
||
midValue = root.left.value
|
||
}
|
||
// return关键字可以省略
|
||
midValue + sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right)
|
||
}
|
||
}
|
||
```
|
||
|
||
**迭代:**
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable
|
||
def sumOfLeftLeaves(root: TreeNode): Int = {
|
||
val stack = mutable.Stack[TreeNode]()
|
||
if (root == null) return 0
|
||
stack.push(root)
|
||
var sum = 0
|
||
while (!stack.isEmpty) {
|
||
val curNode = stack.pop()
|
||
if (curNode.left != null && curNode.left.left == null && curNode.left.right == null) {
|
||
sum += curNode.left.value // 如果满足条件就累加
|
||
}
|
||
if (curNode.right != null) stack.push(curNode.right)
|
||
if (curNode.left != null) stack.push(curNode.left)
|
||
}
|
||
sum
|
||
}
|
||
}
|
||
```
|
||
|
||
### Rust:
|
||
|
||
**递归**
|
||
|
||
```rust
|
||
use std::cell::RefCell;
|
||
use std::rc::Rc;
|
||
impl Solution {
|
||
pub fn sum_of_left_leaves(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
|
||
let mut res = 0;
|
||
if let Some(node) = root {
|
||
if let Some(left) = &node.borrow().left {
|
||
if left.borrow().right.is_none() && left.borrow().right.is_none() {
|
||
res += left.borrow().val;
|
||
}
|
||
}
|
||
res + Self::sum_of_left_leaves(node.borrow().left.clone())
|
||
+ Self::sum_of_left_leaves(node.borrow().right.clone())
|
||
} else {
|
||
0
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
**迭代：**
|
||
|
||
```rust
|
||
use std::cell::RefCell;
|
||
use std::rc::Rc;
|
||
impl Solution {
|
||
pub fn sum_of_left_leaves(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
|
||
let mut res = 0;
|
||
let mut stack = vec![root];
|
||
while !stack.is_empty() {
|
||
if let Some(node) = stack.pop().unwrap() {
|
||
if let Some(left) = &node.borrow().left {
|
||
if left.borrow().left.is_none() && left.borrow().right.is_none() {
|
||
res += left.borrow().val;
|
||
}
|
||
stack.push(Some(left.to_owned()));
|
||
}
|
||
if let Some(right) = &node.borrow().right {
|
||
stack.push(Some(right.to_owned()));
|
||
}
|
||
}
|
||
}
|
||
res
|
||
}
|
||
}
|
||
```
|
||
### C#
|
||
```csharp
|
||
// 递归
|
||
public int SumOfLeftLeaves(TreeNode root)
|
||
{
|
||
if (root == null) return 0;
|
||
|
||
int leftValue = SumOfLeftLeaves(root.left);
|
||
if (root.left != null && root.left.left == null && root.left.right == null)
|
||
{
|
||
leftValue += root.left.val;
|
||
}
|
||
int rightValue = SumOfLeftLeaves(root.right);
|
||
return leftValue + rightValue;
|
||
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|
||
|