LeetCode226. Invert Binary Tree

Invert Binary Tree"/>

LeetCode226. Invert Binary Tree

文章目录

• • 一、题目
  • 二、题解
  • • 2.1 前序遍历版本
    • 2.2 中序遍历版本
    • 2.3 后序遍历版本

一、题目

Given the root of a binary tree, invert the tree, and return its root.

Example 1:

Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:

Input: root = [2,1,3]
Output: [2,3,1]
Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

二、题解

2.1 前序遍历版本

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if(root == nullptr) return root;swap(root->left,root->right);invertTree(root->left);invertTree(root->right);return root;}
};

2.2 中序遍历版本

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if(root == nullptr) return root;invertTree(root->left);swap(root->left,root->right);invertTree(root->left);return root;}
};

2.3 后序遍历版本

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if(root == nullptr) return root;invertTree(root->left);invertTree(root->right);swap(root->left,root->right);return root;}
};