381 lines
11 KiB
Markdown
381 lines
11 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.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.com/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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# 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：**如果左节点不为空，且左节点没有左右孩子，那么这个节点就是左叶子**
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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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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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```
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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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依然是
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```
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if (root == NULL) return 0;
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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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int rightValue = sumOfLeftLeaves(root->right); // 右
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// 中
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int midValue = 0;
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if (root->left && !root->left->left && !root->left->right) {
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midValue = root->left->val;
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}
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int sum = midValue + leftValue + rightValue;
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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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int leftValue = sumOfLeftLeaves(root->left); // 左
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int rightValue = sumOfLeftLeaves(root->right); // 右
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// 中
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int midValue = 0;
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if (root->left && !root->left->left && !root->left->right) { // 中
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midValue = root->left->val;
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}
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int sum = midValue + leftValue + rightValue;
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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 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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return midValue + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
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}
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};
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```
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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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**递归后序遍历**
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```python3
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class Solution:
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def sumOfLeftLeaves(self, root: TreeNode) -> int:
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if not root:
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return 0
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left_left_leaves_sum = self.sumOfLeftLeaves(root.left) # 左
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right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右
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cur_left_leaf_val = 0
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if root.left and not root.left.left and not root.left.right:
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cur_left_leaf_val = root.left.val
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return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中
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```
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**迭代**
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```python3
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class Solution:
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def sumOfLeftLeaves(self, root: TreeNode) -> int:
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"""
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Idea: Each time check current node's left node.
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If current node don't have one, skip it.
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"""
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stack = []
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if root:
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stack.append(root)
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res = 0
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while stack:
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# 每次都把当前节点的左节点加进去.
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cur_node = stack.pop()
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if cur_node.left and not cur_node.left.left and not cur_node.left.right:
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res += cur_node.left.val
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if cur_node.left:
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stack.append(cur_node.left)
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if cur_node.right:
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stack.append(cur_node.right)
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return res
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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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var res int
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findLeft(root,&res)
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return res
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}
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func findLeft(root *TreeNode,res *int){
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//左节点
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if root.Left!=nil&&root.Left.Left==nil&&root.Left.Right==nil{
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*res=*res+root.Left.Val
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}
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if root.Left!=nil{
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findLeft(root.Left,res)
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}
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if root.Right!=nil{
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findLeft(root.Right,res)
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}
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}
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```
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**迭代法**
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```go
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func sumOfLeftLeaves(root *TreeNode) int {
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var res int
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queue:=list.New()
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queue.PushBack(root)
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for queue.Len()>0{
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length:=queue.Len()
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for i:=0;i<length;i++{
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node:=queue.Remove(queue.Front()).(*TreeNode)
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if node.Left!=nil&&node.Left.Left==nil&&node.Left.Right==nil{
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res=res+node.Left.Val
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}
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if node.Left!=nil{
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queue.PushBack(node.Left)
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}
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if node.Right!=nil{
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queue.PushBack(node.Right)
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}
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}
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}
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return res
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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 = sumOfLeftLeaves(node.left);
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let rightValue = sumOfLeftLeaves(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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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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