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 [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \3 4 4 3But the following [1,2,2,null,3,null,3] is not: 1 / \ 2 2 \ \ 3 3Note:Bonus points if you could solve it both recursively and iteratively.难度：easy题目：给定二叉树检查它是否为镜像。思路：同same TreeRuntime: 7 ms, faster than 46.76% of Java online submissions for Symmetric Tree.Memory Usage: 30.1 MB, less than 0.94% of Java online submissions for Symmetric Tree./** * 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 (null == root) { return true; } return isSymmetric(root.left, root.right); } public boolean isSymmetric(TreeNode p, TreeNode q) { if (p == null && q != null || p != null && q == null) { return false; } if (p == null && q == null) { return true; } if (p.val != q.val) { return false; } return isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left); }}