一.解法

https://leetcode-cn.com/problems/invert-binary-tree/
要点：递归，树
Python，C++，Java都用了二叉树递归的方法，如果当前节点是空就返回空，当前节点有值，就交换左孩子和右孩子后返回当前节点。

二.Python实现

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        temp=TreeNode(0)
        if root==None:
            return None
        else:
            temp=root.left
            root.left=root.right
            root.right=temp
            if root.left!=None:
                self.invertTree(root.left)
            if root.right!=None:
                self.invertTree(root.right)
        return root

三.C++实现

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

四.java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        TreeNode temp;
        if(root==null) return null;
        else{
            temp=root.left;
            root.left=root.right;
            root.right=temp;
            if(root.left!=null) invertTree(root.left);
            if(root.right!=null) invertTree(root.right);
        } 
        return root;

    }
}