101. 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.
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"{1,#,2,3}"

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dfs算法

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private boolean isSym (TreeNode left,TreeNode right){
if(left==null&&right==null)return true;
if(left==null||right==null)return false;
if(left.val!=right.val)return false;
return (isSym(left.left,right.right)&&isSym(left.right,right.left));
}
public boolean isSymmetric(TreeNode root) {
if(root==null)return true;
return isSym(root.left,root.right);
}
}