【力扣热题100】[Java版] 刷题笔记-101. 对称二叉树

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);
    }
}