题目

给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树[1,2,2,3,4,4,3]是对称的。

    1
   / \
  2   2
 / \ / \
3  4 4  3

但是下面这个[1,2,2,null,3,null,3]则不是镜像对称的:

    1
   / \
  2   2
   \   \
   3    3

解法

解法一： 递归。

/**
 * 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 {
    
    // 方法一：递归
    public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        TreeNode left = root.left;
        TreeNode right = root.right;
        return isSymmetricForLeftAndRight(left,right);
    }

    // 判断left和right是否对称
    public boolean isSymmetricForLeftAndRight(TreeNode left,TreeNode right){
        if(left == null || right == null){
            if(left != null || right != null){
                return false;
            } else {
                return true;
            }
        }
        // 判断值是否相等
        int leftVal = left.val;
        int rightVal = right.val;
        if(leftVal != rightVal){
            return false;
        }

        // 递归调用，判断左节点的左节点与右节点的右节点是否对称
        TreeNode leftNextLeft = left.left;
        TreeNode rightNextRight = right.right;
        if(!isSymmetricForLeftAndRight(leftNextLeft,rightNextRight)){
            return false;
        }

        // 递归调用，判断左节点的右节点与右节点的左节点是否对称
        TreeNode leftNextRight = left.right;
        TreeNode rightNextLeft = right.left;
        if(!isSymmetricForLeftAndRight(leftNextRight,rightNextLeft)){
            return false;
        }

        return true;
    }
}

解法二： 迭代。

/**
 * 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 {
  
    // 解法二：迭代
    public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        // 创建队列，特点是先进先出。
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);

        while(!queue.isEmpty()){
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            if(left == null || right == null){
                if(left != null || right != null){
                    return false;
                } else {
                    continue;
                }
            }
            int leftVal = left.val;
            int rightVal = right.val;
            if(leftVal != rightVal){
                return false;
            }
            
            // 再把left和right子节点放进去，注意顺序
            // 顺序一定是left的左节点和right的右节点，然后是left的右节点和right的左节点
            queue.offer(left.left);
            queue.offer(right.right);
            queue.offer(left.right);
            queue.offer(right.left);
        }

        return true;
    }
}

总结

本篇文章讲解了算法题目的思路和解法，代码和笔记由于纯手打，难免会有纰漏，如果发现错误的地方，请第一时间告诉我，这将是我进步的一个很重要的环节。以后会定期更新算法题目以及各种开发知识点，如果您觉得写得不错，不妨点个关注，谢谢。