LeetCode 226 Invert Binary Tree

Problem:

Invert a binary tree.

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

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

Analysis:

1、helper function + recursion

一开始我用调用子函数，在子函数中递归的方式实现：

 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 private:
12     void swapNode (TreeNode* root) {
13         TreeNode *tmp = root->left;
14         root->left = root->right;
15         root->right = tmp;
16         
17         if (root->left != NULL) {
18             swapNode (root->left);
19         }
20         
21         if (root->right != NULL) {
22             swapNode (root->right);
23         }
24     }
25     
26 public:
27     TreeNode* invertTree(TreeNode* root) {
28         if (root != NULL) {
29             swapNode (root);
30         }
31         
32         return root;
33     }
34 };

虽然解决了问题，但效率太低

2、recursion

进而将递归写入主函数中，并简化了部分代码

 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) {
14             return root;
15         }
16         
17         TreeNode *tmp = root->left;
18         root->left = invertTree(root->right);
19         root->right = invertTree(tmp);
20         
21         return root;
22     }
23 };

效率有所提高

3、非递归方法

 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) {
14             return NULL;
15         }
16         
17         queue<TreeNode*> q;
18         q.push(root);
19         
20         while (!q.empty()) {
21             TreeNode* cur = q.front();
22             q.pop();
23             
24             TreeNode* tmp = cur->left;
25             cur->left = cur->right;
26             cur->right = tmp;
27             
28             if (cur->left != NULL) {
29                 q.push(cur->left);
30             }
31             
32             if (cur->right != NULL) {
33                 q.push(cur->right);
34             }
35         }
36         
37         return root;
38     }
39 };

来源：https://www.cnblogs.com/VickyWang/p/5992484.html

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left函数