【Leetcode】Symmetric Tree

最新推荐文章于 2025-09-09 20:07:09 发布

转载最新推荐文章于 2025-09-09 20:07:09 发布 · 141 阅读

文章标签：

#leetcode #java #TreeNode #LinkedList

本文介绍了如何通过递归和迭代两种方式检查二叉树是否是对称的，提供了详细的Java代码实现，并解释了二叉树序列化的方法。

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

For example, this binary tree is symmetric:

But the following is not:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

Java:

Recursive:

/**
 * Definition for binary tree
 * 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;
        return isSymmetric(root.left,root.right);
    }
    public boolean isSymmetric(TreeNode a,TreeNode b)
    {
        if(a==null&&b==null) return true;
        if(a!=null&&b==null||a==null&&b!=null) return false;
        if(a.val!=b.val) return false;
        return isSymmetric(a.left,b.right)&&isSymmetric(a.right,b.left);
    }
}

Iteration:

public class Solution {
   
   public boolean isSymmetric(TreeNode root) {  
       if(root == null||root.left == null && root.right == null)  
        return true;  
    if(root.left == null || root.right == null)  
        return false;  
    LinkedList<TreeNode> p1 = new LinkedList<TreeNode>();  
    LinkedList<TreeNode> p2 = new LinkedList<TreeNode>();  
    p1.add(root.left);  
    p2.add(root.right);  
    while(!p1.isEmpty()&&!p2.isEmpty())
    {
        TreeNode t1=p1.poll();
        TreeNode t2=p2.poll();
        if(t1.val!=t2.val) return false;
        if(t1.left!=null&&t2.right==null||t1.left==null&&t2.right!=null) return false;
        if(t1.right!=null&&t2.left==null||t1.right==null&&t2.left!=null) return false;
        if(t1.left!=null&&t2.right!=null)
        {
            p1.add(t1.left);
            p2.add(t2.right);
        }
        if(t1.right!=null&&t2.left!=null)
        {
            p1.add(t1.right);
            p2.add(t2.left);
        }
        
        
    }
    return true;
   }
}

Reference:

1. http://blog.youkuaiyun.com/likecool21/article/details/34429071 一个链表

2. http://blog.youkuaiyun.com/linhuanmars/article/details/23072829 这个思路更清晰

3. http://gongxuns.blogspot.com/2012/12/leetcodesymmetric-tree.html