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.
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/** * Definition for binary tree * 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); } private boolean isSymmetric(TreeNode node1, TreeNode node2) { if(node1 == null && node2 != null) return false; if(node1 != null && node2 == null) return false; if(node1 == null && node2 == null) return true; if(node1.val != node2.val) return false; return isSymmetric(node1.left, node2.right) && isSymmetric(node1.right, node2.left); } }
原文：http://www.cnblogs.com/neweracoding/p/5274467.html