Leetcode 226 Invert Binary Tree

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia: This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.


题目大意：


翻转一棵二叉树。


测试样例见题目描述。


花絮：


这道题目的灵感来自于Max Howell的推特原文：


Google：我们90%的工程师在用你写的软件（Homebrew），但你竟然不会在白板上翻转一棵二叉树，所以滚吧。


解题思路：


递归或者队列迭代均可。


递归：


1、交换根节点的左右子树。


2、对左右子树分别执行递归反转 。




 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         //TreeNode* p = root;
14         if(root == NULL)
15             return root;
16         TreeNode* temp = root->left;
17         root->left = root->right;
18         root->right = temp;
19         invertTree(root->left);
20         invertTree(root->right);
21         return root;
22     }
23 };




 




1    TreeNode* invertTree(TreeNode* root) {
2     if(root==NULL)
3             return NULL;
4     TreeNode * ptmpNode = root->left;
5     root->left = invertTree(root->right);
6     root->right = invertTree(ptmpNode);
7     return root;
8     }




 


非递归算法：


1、交换根节点的左右子节点


2、交换第二层每个节点的左右子节点


....


这个与二叉树层次遍历类似，代码如下：




 1     TreeNode* invertTree(TreeNode* root) {
 2         if (root == NULL)
 3             return root;
 4         queue<TreeNode*> tree_queue;
 5         tree_queue.push(root);
 6 
 7         while (!tree_queue.empty()){
 8             TreeNode * pNode = tree_queue.front();
 9             tree_queue.pop();
10 
11             TreeNode * pLeft = pNode->left;
12             pNode->left = pNode->right;
13             pNode->right = pLeft;
14 
15             if (pNode->left)
16                 tree_queue.push(pNode->left);
17             if (pNode->right)
18                 tree_queue.push(pNode->right);
19         }
20         return root;
21     }




 







		

		
posted @ 
2016-07-26 14:53 
琴影 
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