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。例如下所示

      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：二叉树镜像，即将所有节点的左右子孩子交换即可，但是要注意几个特例：1、根节点为NULL的情况；2、当前节点没有左右孩子；下面是递归和非递归实现：

 1 struct TreeNode {
 2     int val;
 3     struct TreeNode *left;
 4     struct TreeNode *right;
 5     TreeNode(int x) :
 6             val(x), left(NULL), right(NULL) {
 7     }
 8 };
 9 
10 
11 void Mirror(TreeNode *pRoot) {
12         if(pRoot == NULL)
13             return;
14         if(pRoot->left==NULL && pRoot->right==NULL)
15             return;
16         TreeNode* temp = NULL;
17         temp = pRoot->right;
18         pRoot->right = pRoot->left;
19         pRoot->left = temp;
20         if(pRoot->left)
21             Mirror(pRoot->left);
22         if(pRoot->right)
23             Mirror(pRoot->right);
24  }

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