力扣热题100题-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。

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？

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/symmetric-tree

解法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;
        }
        return compare(root.left,root.right);
    }

    public boolean compare(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;
        }
        boolean leftCompare = compare(left.right,right.left);
        if(!leftCompare){
            return false;
        }
        boolean rightCompare = compare(left.left,right.right);
        if(!rightCompare){
            return false;
        }
        return true;
    }
}

递归：时间复杂度：O(n) 空间复杂度：O(height)

使用迭代的方法：

/**
 * 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;
        }
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root.left);
        queue.add(root.right);
        while(!queue.isEmpty()){
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();

            //判断left和right是否相同
            if(left == null && right == null){
                continue;
            }

            if(left == null || right == null){
                return false;
            }
            if(left.val != right.val){
                return false;
            }

            queue.add(left.left);
            queue.add(right.right);
            queue.add(left.right);
            queue.add(right.left);
        }
        return queue.isEmpty();
    }

    
}