LeetCode 101. Symmetric Tree

Posted 2022-12-15 月牙儿June

分类 ：Tree; BFS; DFS 

问题描述：

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

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

Note:
Bonus points if you could solve it both recursively and iteratively.

方法思路：

方法一：循环方法

利用两个先进先出队列分别存储根节点的左右两边的子节点

预备知识：

Queue 用法：offer() 进队列；poll()出队列；isEmpty()判断是否为空

Queue为抽象类，LinkedList实现了Queue接口，所以定义队列是使用

 Queue<TreeNode> left=new LinkedList<>();

Java代码示例：

/**
 * Definition for a binary tree node.
 * public class TreeNode 
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x)  val = x; 
 * 
 */
public class Solution 
    public boolean isSymmetric(TreeNode root) 
        if(root == null) return true;
        Queue<TreeNode> left=new LinkedList<>();
        Queue<TreeNode> right=new LinkedList<>();
        left.offer(root);
        right.offer(root);
        while((!left.isEmpty()) && (!right.isEmpty()))
           TreeNode pl=left.poll();
           TreeNode pr=right.poll();
           if(pl.val!=pr.val) return false;
           if(pl.left!=null && pr.right!=null)
               left.offer(pl.left);
               right.offer(pr.right);
           else if((pl.left==null && pr.right!=null) || (pl.left!=null && pr.right==null)) return false;
           if(pl.right!=null && pr.left!=null)
               left.offer(pl.right);
               right.offer(pr.left);
           else if((pl.right!=null && pr.left==null) || (pl.right==null && pr.left!=null))return false;
        
        if((left.isEmpty()) && (right.isEmpty())) return true;
        return false;
    

方法二：递归方法

待续。。。。。。