剑指offer 二叉树的镜像

题目：操作给定的二叉树，将其变换为源二叉树的镜像。

如：

二叉树的镜像定义：源二叉树 
    	    8
    	   /  
    	  6   10
    	 /   / 
    	5  7 9 11
    	镜像二叉树
    	    8
    	   /  
    	  10   6
    	 /   / 
    	11 9 7  5

分析：递归的交换左右子树。也可以用栈实现。

解法一：递归

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     void Mirror(TreeNode *pRoot) {
13         if (pRoot == nullptr) {
14             return;
15         }
16         TreeNode *tmp = pRoot->left;
17         pRoot->left = pRoot->right;
18         pRoot->right = tmp;
19         Mirror(pRoot->left);
20         Mirror(pRoot->right);
21     }
22 };

解法二：栈实现

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     void Mirror(TreeNode *pRoot) {
13         if (pRoot == NULL)
14             return;
15         stack<TreeNode*> st;
16         st.push(pRoot);
17         while (!st.empty()) {
18             TreeNode* top = st.top(), *tmp;
19             st.pop();
20             tmp = top->left;
21             top->left = top->right;
22             top->right = tmp;
23             if (top->left != NULL)
24                 st.push(top->left);
25             if (top->right != NULL)
26                 st.push(top->right);
27         }
28     }
29 };