572 lines
15 KiB
Markdown
572 lines
15 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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# 二叉树的递归遍历
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《代码随想录》算法视频公开课：[每次写递归都要靠直觉？ 这次带你学透二叉树的递归遍历！](https://www.bilibili.com/video/BV1Wh411S7xt)，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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> 一看就会，一写就废！
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这次我们要好好谈一谈递归，为什么很多同学看递归算法都是“一看就会，一写就废”。
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主要是对递归不成体系，没有方法论，**每次写递归算法 ，都是靠玄学来写代码**，代码能不能编过都靠运气。
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**本篇将介绍前后中序的递归写法，一些同学可能会感觉很简单，其实不然，我们要通过简单题目把方法论确定下来，有了方法论，后面才能应付复杂的递归。**
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这里帮助大家确定下来递归算法的三个要素。**每次写递归，都按照这三要素来写，可以保证大家写出正确的递归算法！**
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1. **确定递归函数的参数和返回值：**
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确定哪些参数是递归的过程中需要处理的，那么就在递归函数里加上这个参数， 并且还要明确每次递归的返回值是什么进而确定递归函数的返回类型。
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2. **确定终止条件：**
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写完了递归算法, 运行的时候，经常会遇到栈溢出的错误，就是没写终止条件或者终止条件写的不对，操作系统也是用一个栈的结构来保存每一层递归的信息，如果递归没有终止，操作系统的内存栈必然就会溢出。
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3. **确定单层递归的逻辑：**
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确定每一层递归需要处理的信息。在这里也就会重复调用自己来实现递归的过程。
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好了，我们确认了递归的三要素，接下来就来练练手：
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**以下以前序遍历为例：**
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1. **确定递归函数的参数和返回值**：因为要打印出前序遍历节点的数值，所以参数里需要传入vector来放节点的数值，除了这一点就不需要再处理什么数据了也不需要有返回值，所以递归函数返回类型就是void，代码如下：
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```cpp
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void traversal(TreeNode* cur, vector<int>& vec)
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```
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2. **确定终止条件**：在递归的过程中，如何算是递归结束了呢，当然是当前遍历的节点是空了，那么本层递归就要结束了，所以如果当前遍历的这个节点是空，就直接return，代码如下：
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```cpp
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if (cur == NULL) return;
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```
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3. **确定单层递归的逻辑**：前序遍历是中左右的循序，所以在单层递归的逻辑，是要先取中节点的数值，代码如下：
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```cpp
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vec.push_back(cur->val); // 中
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traversal(cur->left, vec); // 左
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traversal(cur->right, vec); // 右
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```
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单层递归的逻辑就是按照中左右的顺序来处理的，这样二叉树的前序遍历，基本就写完了，再看一下完整代码：
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前序遍历：
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```CPP
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class Solution {
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public:
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void traversal(TreeNode* cur, vector<int>& vec) {
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if (cur == NULL) return;
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vec.push_back(cur->val); // 中
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traversal(cur->left, vec); // 左
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traversal(cur->right, vec); // 右
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}
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vector<int> preorderTraversal(TreeNode* root) {
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vector<int> result;
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traversal(root, result);
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return result;
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}
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};
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```
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那么前序遍历写出来之后，中序和后序遍历就不难理解了，代码如下：
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中序遍历：
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```CPP
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void traversal(TreeNode* cur, vector<int>& vec) {
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if (cur == NULL) return;
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traversal(cur->left, vec); // 左
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vec.push_back(cur->val); // 中
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traversal(cur->right, vec); // 右
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}
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```
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后序遍历：
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```CPP
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void traversal(TreeNode* cur, vector<int>& vec) {
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if (cur == NULL) return;
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traversal(cur->left, vec); // 左
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traversal(cur->right, vec); // 右
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vec.push_back(cur->val); // 中
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}
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```
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此时大家可以做一做leetcode上三道题目，分别是：
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* [144.二叉树的前序遍历](https://leetcode.cn/problems/binary-tree-preorder-traversal/)
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* [145.二叉树的后序遍历](https://leetcode.cn/problems/binary-tree-postorder-traversal/)
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* [94.二叉树的中序遍历](https://leetcode.cn/problems/binary-tree-inorder-traversal/)
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可能有同学感觉前后中序遍历的递归太简单了，要打迭代法（非递归），别急，我们明天打迭代法，打个通透！
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# 其他语言版本
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Java：
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```Java
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// 前序遍历·递归·LC144_二叉树的前序遍历
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class Solution {
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public List<Integer> preorderTraversal(TreeNode root) {
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List<Integer> result = new ArrayList<Integer>();
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preorder(root, result);
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return result;
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}
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public void preorder(TreeNode root, List<Integer> result) {
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if (root == null) {
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return;
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}
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result.add(root.val);
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preorder(root.left, result);
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preorder(root.right, result);
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}
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}
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// 中序遍历·递归·LC94_二叉树的中序遍历
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class Solution {
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public List<Integer> inorderTraversal(TreeNode root) {
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List<Integer> res = new ArrayList<>();
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inorder(root, res);
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return res;
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}
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void inorder(TreeNode root, List<Integer> list) {
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if (root == null) {
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return;
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}
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inorder(root.left, list);
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list.add(root.val); // 注意这一句
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inorder(root.right, list);
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}
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}
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// 后序遍历·递归·LC145_二叉树的后序遍历
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class Solution {
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public List<Integer> postorderTraversal(TreeNode root) {
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List<Integer> res = new ArrayList<>();
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postorder(root, res);
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return res;
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}
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void postorder(TreeNode root, List<Integer> list) {
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if (root == null) {
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return;
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}
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postorder(root.left, list);
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postorder(root.right, list);
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list.add(root.val); // 注意这一句
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}
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}
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```
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Python：
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```python
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# 前序遍历-递归-LC144_二叉树的前序遍历
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class Solution:
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def preorderTraversal(self, root: TreeNode) -> List[int]:
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# 保存结果
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result = []
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def traversal(root: TreeNode):
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if root == None:
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return
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result.append(root.val) # 前序
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traversal(root.left) # 左
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traversal(root.right) # 右
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traversal(root)
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return result
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# 中序遍历-递归-LC94_二叉树的中序遍历
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class Solution:
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def inorderTraversal(self, root: TreeNode) -> List[int]:
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result = []
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def traversal(root: TreeNode):
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if root == None:
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return
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traversal(root.left) # 左
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result.append(root.val) # 中序
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traversal(root.right) # 右
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traversal(root)
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return result
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# 后序遍历-递归-LC145_二叉树的后序遍历
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class Solution:
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def postorderTraversal(self, root: TreeNode) -> List[int]:
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result = []
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def traversal(root: TreeNode):
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if root == None:
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return
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traversal(root.left) # 左
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traversal(root.right) # 右
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result.append(root.val) # 后序
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traversal(root)
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return result
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```
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Go：
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前序遍历:
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```go
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func preorderTraversal(root *TreeNode) (res []int) {
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var traversal func(node *TreeNode)
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traversal = func(node *TreeNode) {
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if node == nil {
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return
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}
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res = append(res,node.Val)
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traversal(node.Left)
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traversal(node.Right)
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}
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traversal(root)
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return res
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}
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```
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中序遍历:
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```go
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func inorderTraversal(root *TreeNode) (res []int) {
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var traversal func(node *TreeNode)
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traversal = func(node *TreeNode) {
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if node == nil {
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return
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}
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traversal(node.Left)
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res = append(res,node.Val)
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traversal(node.Right)
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}
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traversal(root)
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return res
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}
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```
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后序遍历:
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```go
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func postorderTraversal(root *TreeNode) (res []int) {
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var traversal func(node *TreeNode)
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traversal = func(node *TreeNode) {
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if node == nil {
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return
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}
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traversal(node.Left)
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traversal(node.Right)
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res = append(res,node.Val)
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}
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traversal(root)
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return res
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}
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```
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Javascript版本：
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前序遍历：
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```Javascript
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var preorderTraversal = function(root) {
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let res=[];
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const dfs=function(root){
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if(root===null)return ;
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//先序遍历所以从父节点开始
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res.push(root.val);
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//递归左子树
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dfs(root.left);
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//递归右子树
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dfs(root.right);
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}
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//只使用一个参数 使用闭包进行存储结果
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dfs(root);
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return res;
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};
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```
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中序遍历
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```javascript
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var inorderTraversal = function(root) {
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let res=[];
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const dfs=function(root){
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if(root===null){
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return ;
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}
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dfs(root.left);
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res.push(root.val);
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dfs(root.right);
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}
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dfs(root);
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return res;
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};
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```
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后序遍历
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```javascript
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var postorderTraversal = function(root) {
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let res=[];
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const dfs=function(root){
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if(root===null){
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return ;
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}
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dfs(root.left);
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dfs(root.right);
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res.push(root.val);
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}
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dfs(root);
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return res;
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};
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```
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TypeScript:
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```typescript
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// 前序遍历
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function preorderTraversal(node: TreeNode | null): number[] {
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function traverse(node: TreeNode | null, res: number[]): void {
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if (node === null) return;
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res.push(node.val);
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traverse(node.left, res);
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traverse(node.right, res);
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}
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const res: number[] = [];
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traverse(node, res);
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return res;
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}
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// 中序遍历
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function inorderTraversal(node: TreeNode | null): number[] {
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function traverse(node: TreeNode | null, res: number[]): void {
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if (node === null) return;
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traverse(node.left, res);
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res.push(node.val);
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traverse(node.right, res);
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}
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const res: number[] = [];
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traverse(node, res);
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return res;
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}
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// 后序遍历
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function postorderTraversal(node: TreeNode | null): number[] {
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function traverse(node: TreeNode | null, res: number[]): void {
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if (node === null) return;
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traverse(node.left, res);
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traverse(node.right, res);
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res.push(node.val);
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}
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const res: number[] = [];
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traverse(node, res);
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return res;
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}
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```
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C:
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```c
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//前序遍历：
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void preOrder(struct TreeNode* root, int* ret, int* returnSize) {
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if(root == NULL)
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return;
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ret[(*returnSize)++] = root->val;
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preOrder(root->left, ret, returnSize);
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preOrder(root->right, ret, returnSize);
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}
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int* preorderTraversal(struct TreeNode* root, int* returnSize){
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int* ret = (int*)malloc(sizeof(int) * 100);
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*returnSize = 0;
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preOrder(root, ret, returnSize);
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return ret;
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}
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//中序遍历：
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void inOrder(struct TreeNode* node, int* ret, int* returnSize) {
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if(!node)
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return;
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inOrder(node->left, ret, returnSize);
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ret[(*returnSize)++] = node->val;
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inOrder(node->right, ret, returnSize);
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}
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int* inorderTraversal(struct TreeNode* root, int* returnSize){
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int* ret = (int*)malloc(sizeof(int) * 100);
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*returnSize = 0;
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inOrder(root, ret, returnSize);
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return ret;
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}
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//后序遍历：
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void postOrder(struct TreeNode* node, int* ret, int* returnSize) {
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if(node == NULL)
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return;
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postOrder(node->left, ret, returnSize);
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postOrder(node->right, ret, returnSize);
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ret[(*returnSize)++] = node->val;
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}
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int* postorderTraversal(struct TreeNode* root, int* returnSize){
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int* ret= (int*)malloc(sizeof(int) * 100);
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*returnSize = 0;
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postOrder(root, ret, returnSize);
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return ret;
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}
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```
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Swift:
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前序遍历：（144.二叉树的前序遍历）
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```Swift
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func preorderTraversal(_ root: TreeNode?) -> [Int] {
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var res = [Int]()
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preorder(root, res: &res)
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return res
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}
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func preorder(_ root: TreeNode?, res: inout [Int]) {
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if root == nil {
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return
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}
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res.append(root!.val)
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preorder(root!.left, res: &res)
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preorder(root!.right, res: &res)
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}
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```
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中序遍历：（94. 二叉树的中序遍历）
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```Swift
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func inorderTraversal(_ root: TreeNode?) -> [Int] {
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var res = [Int]()
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inorder(root, res: &res)
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return res
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}
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func inorder(_ root: TreeNode?, res: inout [Int]) {
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if root == nil {
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return
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}
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inorder(root!.left, res: &res)
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res.append(root!.val)
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inorder(root!.right, res: &res)
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}
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```
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后序遍历：（145. 二叉树的后序遍历）
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```Swift
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func postorderTraversal(_ root: TreeNode?) -> [Int] {
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var res = [Int]()
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postorder(root, res: &res)
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return res
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}
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func postorder(_ root: TreeNode?, res: inout [Int]) {
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if root == nil {
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return
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}
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postorder(root!.left, res: &res)
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postorder(root!.right, res: &res)
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res.append(root!.val)
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}
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```
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Scala: 前序遍历：（144.二叉树的前序遍历）
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```scala
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object Solution {
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import scala.collection.mutable.ListBuffer
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def preorderTraversal(root: TreeNode): List[Int] = {
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val res = ListBuffer[Int]()
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def traversal(curNode: TreeNode): Unit = {
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if(curNode == null) return
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res.append(curNode.value)
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traversal(curNode.left)
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traversal(curNode.right)
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}
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traversal(root)
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res.toList
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}
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}
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```
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中序遍历：（94. 二叉树的中序遍历）
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```scala
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object Solution {
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import scala.collection.mutable.ListBuffer
|
||
def inorderTraversal(root: TreeNode): List[Int] = {
|
||
val res = ListBuffer[Int]()
|
||
def traversal(curNode: TreeNode): Unit = {
|
||
if(curNode == null) return
|
||
traversal(curNode.left)
|
||
res.append(curNode.value)
|
||
traversal(curNode.right)
|
||
}
|
||
traversal(root)
|
||
res.toList
|
||
}
|
||
}
|
||
```
|
||
后序遍历：（145. 二叉树的后序遍历）
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable.ListBuffer
|
||
def postorderTraversal(root: TreeNode): List[Int] = {
|
||
val res = ListBuffer[Int]()
|
||
def traversal(curNode: TreeNode): Unit = {
|
||
if (curNode == null) return
|
||
traversal(curNode.left)
|
||
traversal(curNode.right)
|
||
res.append(curNode.value)
|
||
}
|
||
traversal(root)
|
||
res.toList
|
||
}
|
||
}
|
||
```
|
||
|
||
rust:
|
||
|
||
```rust
|
||
use std::cell::RefCell;
|
||
use std::rc::Rc;
|
||
impl Solution {
|
||
pub fn preorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
|
||
let mut res = vec![];
|
||
Self::traverse(&root, &mut res);
|
||
res
|
||
}
|
||
|
||
//前序遍历
|
||
pub fn traverse(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) {
|
||
if let Some(node) = root {
|
||
res.push(node.borrow().val);
|
||
Self::traverse(&node.borrow().left, res);
|
||
Self::traverse(&node.borrow().right, res);
|
||
}
|
||
}
|
||
//后序遍历
|
||
pub fn traverse(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) {
|
||
if let Some(node) = root {
|
||
Self::traverse(&node.borrow().left, res);
|
||
Self::traverse(&node.borrow().right, res);
|
||
res.push(node.borrow().val);
|
||
}
|
||
}
|
||
//中序遍历
|
||
pub fn traverse(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) {
|
||
if let Some(node) = root {
|
||
Self::traverse(&node.borrow().left, res);
|
||
res.push(node.borrow().val);
|
||
Self::traverse(&node.borrow().right, res);
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
<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>
|