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LeetCode: 226 Invert Binary Tree(easy)

题目：

Invert a binary tree.

     4
   /   
  2     7
 /    / 
1   3 6   9

     4
   /   
  7     2
 /    / 
9   6 3   1

代码：

自己的（递归实现）：

 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* tem;
14         if (!root)
15             return 0;             
16         tem = root->left;
17         root->left = root->right;
18         root->right = tem;
19         if(root->left)
20             invertTree(root->left);
21         if(root->right)
22             invertTree(root->right);
23         return root;
24     }
25 };

别人的：

 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){
14             swap(root->left, root->right);
15             invertTree(root->left);
16             invertTree(root->right);     
17         }
18         return root;
19     }
20 };

自己的（使用队列，非递归实现）:

 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         queue<TreeNode*> queue;
14         if(!root)
15             return 0;
16         queue.push(root);
17         TreeNode *curNode;
18         while(!queue.empty()){
19             curNode = queue.front();
20             queue.pop();
21             swap(curNode->left, curNode->right);
22             if(curNode->left)
23                 queue.push(curNode->left);
24             if(curNode->right)
25                 queue.push(curNode->right);        
26         }
27         return root;
28     }
29 };

非递归和递归的在LeetCode上运行效率差不多。。。

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原文地址：https://www.cnblogs.com/llxblogs/p/7445551.html