刷题18 二叉树的镜像

Posted 2020-10-19 purehol

描述：  操作给定的二叉树，将其变换为源二叉树的镜像。

二叉树的镜像示例：

　　　　   源二叉树 
    	    8
    	   /      	  6   10
    	 / \  /     	5  7 9 11
    	镜像二叉树
    	    8
    	   /      	  10   6
    	 / \  /     	11 9 7  5

很容易的想到了递归：

 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)
14             return;
15         if(!pRoot->left && !pRoot->right)
16             return;
17         TreeNode *pTemp = pRoot->left;
18         pRoot->left = pRoot->right;
19         pRoot->right = pTemp;
20         Mirror(pRoot->left);
21         Mirror(pRoot->right);
22     }
23 };

 

接下来再用非递归栈实现一次：

 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)
14             return;
15         stack<TreeNode*> pStack;
16         pStack.push(pRoot);
17         while(!pStack.empty())
18         {
19             TreeNode *cur = pStack.top();
20             pStack.pop();
21             if(cur->left || cur->right)
22             {
23                 TreeNode *temp = cur->left;
24                 cur->left = cur->right;
25                 cur->right = temp;
26             }
27             if(cur->left)
28                 pStack.push(cur->left);
29             if(cur->right)
30                 pStack.push(cur->right);
31         }
32     }
33 };