题目：101. 对称二叉树

给你一个二叉树的根节点 root ， 检查它是否轴对称。

解题思路

可以理解为遍历对比，最简单的方法就是递归。

解题过程

递归：左右子树分开遍历，左子树遵循根、左、右的顺序，右子树循序根、右、左，将结果进行对比。

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public void leftTravel(TreeNode root, List<Integer> res) {if (root == null) {res.add(null);return;}res.add(root.val);leftTravel(root.left, res);leftTravel(root.right, res);}public void rightTravel(TreeNode root, List<Integer> res) {if (root == null) {res.add(null);return;}res.add(root.val);rightTravel(root.right, res);rightTravel(root.left, res);}public boolean isSymmetric(TreeNode root) {if (root == null) {return true;}if (root.left == null && root.right == null) {return true;}// 左子树 根 左 右TreeNode left = root.left;List<Integer> resL = new ArrayList();leftTravel(left, resL);// 右子树 根 右 左TreeNode right = root.right;List<Integer> resR = new ArrayList();rightTravel(right, resR);// 对比 是否一致return resL.equals(resR);}
}

这个还可以简化，直接递归对比结果，代码如下

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