[剑指Offer] 18.二叉树的镜像

Posted 2020-09-01 NULL

【思路1】递归，所有孩子交换再分别递归左右子树

 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         TreeNode *temp = pRoot->left;
16         pRoot->left = pRoot->right;
17         pRoot->right = temp;
18         Mirror(pRoot->left);
19         Mirror(pRoot->right);
20     }
21 };

【思路2】非递归

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