[LintCode] 翻转二叉树

递归实现：

 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14 public:
15     /**
16      * @param root: a TreeNode, the root of the binary tree
17      * @return: nothing
18      */
19     void invertBinaryTree(TreeNode *root) {
20         // write your code here
21         if (!root) return;
22         swap(root -> left, root -> right);
23         invertBinaryTree(root -> left);
24         invertBinaryTree(root -> right);
25     }
26 };

迭代实现：

 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14 public:
15     /**
16      * @param root: a TreeNode, the root of the binary tree
17      * @return: nothing
18      */
19     void invertBinaryTree(TreeNode *root) {
20         // write your code here
21         if (!root) return;
22         queue<TreeNode*> level;
23         level.push(root);
24         while (!level.empty()) {
25             TreeNode* node = level.front();
26             level.pop();
27             swap(node -> left, node -> right);
28             if (node -> left) level.push(node -> left);
29             if (node -> right) level.push(node -> right);
30         }
31     }
32 };

 

posted @ 2015-06-28 15:26  jianchao-li  阅读(495)  评论(0)  编辑  收藏  举报