标签：
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.
递归解法：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root ==null){
return true;
}
return isSymmetric(root.left, root.right);
}
public boolean isSymmetric(TreeNode l, TreeNode r){
if(l == null && r == null){
return true;
}
else if(l == null || r == null){
return false;
}
if(l.val != r.val){
return false;
}
if(!isSymmetric(l.left, r.right)){
return false;
}
if(!isSymmetric(l.right, r.left)){
return false;
}
return true;
}
}
标签：
原文地址：http://www.cnblogs.com/incrediblechangshuo/p/5464015.html
