LeetCode Symmetric Tree

最新推荐文章于 2025-08-08 21:26:37 发布

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本文详细阐述了如何使用非递归方法判断二叉树是否对称，包括利用双队列模拟遍历过程，并特别注意了处理叶子节点避免空指针错误的情况。同时提供了递归和迭代两种解决方案，帮助读者深入理解二叉树的相关概念。

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题目：

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}".

题意：

就是判断一棵二叉树是否是对称的。

我还是采用非递归的方式，用两个队列LinkedList的方法，将每一棵二叉树的节点入队，然后依次判断。当然需要注意的是，碰到叶子节点的情况，很容易出现空指针错误。所以就需要注意，在if判断的时候要留个心眼儿。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution 
{
    public static boolean isSymmetric(TreeNode root)
	{
		if(root == null || ((root.left == null) && (root.right == null)))
			return true;
		else if((root.left != null && root.right == null ) || (root.left == null && root.right != null ))
			return false;
		else
		{
			LinkedList<TreeNode> nodeleft = new LinkedList<TreeNode>(); //用两个队列来分别遍历树的左右子树
			LinkedList<TreeNode> noderight = new LinkedList<TreeNode>(); 
			nodeleft.add(root);
			noderight.add(root);
			while(!nodeleft.isEmpty() && !noderight.isEmpty())
			{
				TreeNode left = nodeleft.peek();
				TreeNode right = noderight.peek();
				nodeleft.poll();
				noderight.poll();
				if((left.left == null && right.right != null ))
					return false;
				else if(left.left != null && right.right == null )
					return false;
				else if(left.left != null && right.right != null && left.left.val != right.right.val) //这里需要注意叶子节点
					return false;
				else if(left.right == null && right.left != null)
					return false;
				else if(left.right != null && right.left == null)
					return false;
				else if(right.left != null && left.right != null && left.right.val != right.left.val)
					return false;
				else 
				{
					if((left == root) && (right == root))
					{
						if(left.left != null)
							nodeleft.add(left.left);
						if(right.right != null)
							noderight.add(right.right);
					}
					else
					{
						if(left.left != null)
						    nodeleft.add(left.left);
						if(left.right != null)
							nodeleft.add(left.right);
						if(right.right != null)
							noderight.add(right.right);
						if(right.left != null)
							noderight.add(right.left);
					}
				}
			}
			return true;
		}
	}
}

这些题目都是树的典型应用，可以分别和之前做过的一些树的题目相比较，可以发现采用非递归的方法都是用LinkedList来模拟队列，然后让节点入队，依次比较节点之间的值的关系。