leetcode 226. 翻转二叉树

翻转一棵二叉树。

     4
   /     2     7
 / \   / 1   3 6   9

转换为:

     4
   /     7     2
 / \   / 9   6 3   1

注意点:小心不要把程序写成下面这样: 1 root->left = invertTree(root->right);  2 root->right = invert(root->left); 因为第一行的root->left指向的内容已近改变,要用一个变量来保存原来的root->left的值
 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* invertTree(TreeNode* root) {
13         if(root == NULL) return NULL;
14         TreeNode* temp = root->left;
15         root->left = invertTree(root->right);
16         root->right = invertTree(temp);
17         return root;
18     }
19 };

原文地址:https://www.cnblogs.com/mr-stn/p/8978015.html

时间: 2024-10-07 15:50:17

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