34-二叉树练习-LeetCode101对称二叉树-优快云博客

题目

给你一个二叉树的根节点 root ，检查它是否轴对称。

示例 1：

输入：root = [1,2,2,3,4,4,3]
输出：true

示例 2：

输入：root = [1,2,2,null,3,null,3]
输出：false

提示：

树中节点数目在范围 [1, 1000] 内
-100 <= Node.val <= 100

进阶：你可以运用递归和迭代两种方法解决这个问题吗？

思路1：递归

代码1

/**
 * 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 isSymmetric(TreeNode root) {
        if(root == null) {
            return true;
        }
        //原方法isSymmetric只传了1个参数root
        //要判断root.left和root.right是否互为镜像，要再定义一个方法，传入两个参数：左树，右树
        return isMirror(root.left, root.right);
    }

    /**
     * 判断传入的2棵树是否互为镜像关系
     * @param left
     * @param right
     * @return
     */
    public boolean isMirror(TreeNode left, TreeNode right) {
        if(left == null && right == null) {
            return true;
        }
        if(left == null || right == null) {
            return false;
        }
        if(left.val != right.val) {
            return false;
        }
        //以上3个if判断不可以写成root.left和root.right放在上面的isSymmetric()方法中
        //因为如果那样写，只能判断当前节点，左右子树就判断不到了
        
        //判断left.left和right.right是否镜像，left.right和right.left是否镜像
        return isMirror(left.left, right.right) && isMirror(left.right, right.left);
    }
}

思路2：非递归，迭代，队列，层序遍历

代码2

/**
 * 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 isSymmetric(TreeNode root) {
        if(root == null || (root.left == null && root.right == null)) {
            return true;
        }

        //使用队列存储每一层需要判断对称的节点
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);

        while(!queue.isEmpty()) {
            //每次取出一组元素（左树和右树的根）
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();

            //判空
            if(left == null && right == null) {
                continue;
            }
            if(left == null || right == null) {
                return false;
            }
            if(left.val != right.val) {
                return false;
            }
            
            //将当前左右树的根节点判断完毕，将left.left和right.right入队，将left.right和right.left入队
            queue.offer(left.left);
            queue.offer(right.right);
            queue.offer(left.right);
            queue.offer(right.left);
        }
        return true;
    }
}