LeetCode: Symmetric Tree

看了网上答案

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool dfs(TreeNode *left, TreeNode *right) {
13         if (!left && !right) return true;
14         if (!left || !right) return false;
15         return left->val == right->val && dfs(left->left, right->right) && dfs(left->right, right->left);
16     }
17     bool isSymmetric(TreeNode *root) {
18         // Start typing your C/C++ solution below
19         // DO NOT write int main() function
20         if (!root) return true;
21         return dfs(root->left, root->right);
22     }
23 };

下面是自己写的iterasive的法子

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isSymmetric(TreeNode *root) {
13         if (!root) return true;
14         queue<TreeNode*> ltr, rtl;
15         ltr.push(root), rtl.push(root);
16         while (!ltr.empty() && !rtl.empty()) {
17             TreeNode *ltrfront = ltr.front();
18             TreeNode *rtlfront = rtl.front();
19             ltr.pop(), rtl.pop();
20             if (!ltrfront && !rtlfront) continue;
21             if (!ltrfront || !rtlfront || ltrfront->val != rtlfront->val) return false;
22             ltr.push(ltrfront->left), ltr.push(ltrfront->right);
23             rtl.push(rtlfront->right), rtl.push(rtlfront->left);
24         }
25         return ltr.empty() && rtl.empty();
26     }
27 };

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public bool IsSymmetric(TreeNode root) {
12         if (root == null) return true;
13         Queue<TreeNode> ltr = new Queue<TreeNode>();
14         Queue<TreeNode> rtl = new Queue<TreeNode>();
15         ltr.Enqueue(root);
16         rtl.Enqueue(root);
17         while (ltr.Count > 0 && rtl.Count > 0) {
18             TreeNode ltrPeek = ltr.Peek();
19             TreeNode rtlPeek = rtl.Peek();
20             ltr.Dequeue();
21             rtl.Dequeue();
22             if (ltrPeek == null && rtlPeek == null) continue;
23             if (ltrPeek == null || rtlPeek == null || ltrPeek.val != rtlPeek.val) return false;
24             ltr.Enqueue(ltrPeek.left);
25             ltr.Enqueue(ltrPeek.right);
26             rtl.Enqueue(rtlPeek.right);
27             rtl.Enqueue(rtlPeek.left);
28         }
29         return ltr.Count == 0 && rtl.Count == 0;
30     }
31 }

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posted @ 2013-04-22 03:51 ying_vincent 阅读(151) 评论(0) 编辑收藏举报

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