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二叉树镜像

【题目描述】

操作给定的二叉树，将其变换为源二叉树的镜像。 
输入描述:
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
 
【代码实现】
方法1：递归实现
 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|| (pRoot->left==NULL && pRoot->right==NULL))
14             return;
15         TreeNode * node=pRoot->left;
16         pRoot->left=pRoot->right;
17         pRoot->right=node;
18         if(pRoot->left)
19             Mirror(pRoot->left);
20         if(pRoot->right)
21             Mirror(pRoot->right);
22     }
23 };

方法二：迭代实现

 1 /*
 2 struct TreeNode {
 3     int val;//数据
 4     struct TreeNode *left;//左子节点指针
 5     struct TreeNode *right;//右子节点指针
 6     TreeNode(int x) ://用初始化列表进行初始的构造函数
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     void Mirror(TreeNode *pRoot) {
13     if(pRoot==NULL|| (pRoot->left==NULL && pRoot->right==NULL))
14             return;
15         stack<TreeNode*> stackNode;
16         stackNode.push(pRoot);
17         while(stackNode.size()!=0){
18             TreeNode* treeNode=stackNode.top();
19             stackNode.pop();
20             if(treeNode->left!=NULL ||treeNode->right!=NULL){
21                 TreeNode *temp=treeNode->left;
22                treeNode->left=treeNode->right;
23                 treeNode->right=temp;
24             }
25             if(treeNode->left)
26                 stackNode.push(treeNode->left);
27             if(treeNode->right)
28                 stackNode.push(treeNode->right);
29         }
30     }
31 };

 

posted @ 2015-12-06 16:49  GeekerLou  阅读(168)  评论(0)  编辑  收藏  举报