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 3But the following is not: 1 / \ 2 2 \ \ 3 3Note:Bonus points if you could solve it both recursively and iteratively.
难度：82. 这道题是树的题目，本质上还是树的遍历。这里无所谓哪种遍历方式，只需要对相应结点进行比较即可。一颗树对称其实就是看左右子树是否对称，一句话就是左同右，右同左，结点是对称的相等。不对称的条件有以下三个：（1）左边为空而右边不为空；（2）左边不为空而右边为空；（3）左边值不等于右边值。根据这几个条件在遍历时进行判断即可。
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 public class Solution {11 public boolean isSymmetric(TreeNode root) {12 if (root == null) return true;13 return check(root.left, root.right);14 }15 16 public boolean check(TreeNode node1, TreeNode node2) {17 if (node1 == null && node2 == null) return true;18 else if (node1 == null && node2 != null) return false;19 else if (node1 != null && node2 == null) return false;20 else if (node1.val != node2.val) return false;21 return check(node1.left, node2.right) && check(node1.right, node2.left);22 }23 } Iterative的方式参考了网上的解法：
public boolean isSymmetric(TreeNode root) { if(root == null) return true; if(root.left == null && root.right == null) return true; if(root.left == null || root.right == null) return false; LinkedList q1 = new LinkedList (); LinkedList q2 = new LinkedList (); q1.add(root.left); q2.add(root.right); while(!q1.isEmpty() && !q2.isEmpty()) { TreeNode n1 = q1.poll(); TreeNode n2 = q2.poll(); if(n1.val != n2.val) return false; if(n1.left == null && n2.right != null || n1.left != null && n2.right == null) return false; if(n1.right == null && n2.left != null || n1.right != null && n2.left == null) return false; if(n1.left != null && n2.right != null) { q1.add(n1.left); q2.add(n2.right); } if(n1.right != null && n2.left != null) { q1.add(n1.right); q2.add(n2.left); } } return true;}