Leetcode: Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

 

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

 

Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

code:

 1 class Solution {
 2 
 3 public:
 4 
 5 bool isSymmetric(TreeNode *root) {
 6     if(root == NULL) return true;
 7 
 8     queue<TreeNode *> lf, rt;
 9     lf.push(root->left);
10     rt.push(root->right);
11     TreeNode *l, *r;
12     while(!lf.empty() && !rt.empty()) {
13         l = lf.front(); r = rt.front();
14         lf.pop(); rt.pop();
15         if(l == NULL && r == NULL) continue;
16         if(l == NULL || r == NULL) return false;
17         if(l->val != r->val) return false;
18         lf.push(l->left); lf.push(l->right);
19         rt.push(r->right); rt.push(r->left);
20     }
21     if(lf.empty() && rt.empty()) return true;
22     else return false;
23 }
24 };

递归解：

// Recursive solution

public class Solution {

public boolean isSymmetric(TreeNode root) {
    if (root == null) {
        return true;
    }

    return helper(root.left, root.right);
}

private boolean helper(TreeNode a, TreeNode b) {
    if (a == null && b == null) return true;
    if (a == null || b == null) return false;
    if (a.val != b.val) return false;
    return helper(a.left, b.right) && helper(a.right, b.left);
}
}

 

 

posted @ 2013-04-30 13:39  caijinlong  阅读(127)  评论(0)  编辑  收藏  举报