leetcode：Symmetric Tree

最新推荐文章于 2019-08-15 20:43:00 发布

糖小狸

最新推荐文章于 2019-08-15 20:43:00 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/u011235543/article/details/51707427

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本文介绍了一种检查二叉树是否为中心对称的方法，提供了两种实现方案：递归和非递归方式。递归方法通过比较左右子树是否为镜像来判断，而非递归方法使用两个队列来逐层比较节点。

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:

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

/**
 * 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;
        return check(root.left,root.right);
    }
    public boolean check(TreeNode q, TreeNode p)
    {
        if(q==null||p==null)
            return p==q;
        return check(q.left,p.right)&&check(q.right,p.left)&&q.val==p.val;
    }
}

上述方法采用的是递归的思想；另一种不用递归的方法下面给出

方法二：

/**
 * 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) 
    {
        LinkedList<TreeNode> q=new LinkedList<TreeNode>();
        LinkedList<TreeNode> p=new LinkedList<TreeNode>();
        q.addLast(root);
        p.addLast(root);
        while(!q.isEmpty())
        {
            TreeNode t1=q.remove();
            TreeNode t2=p.remove();
  
            if(t1==null&&t2==null)
                continue;
            if(t1==null||t2==null)
                return false;
            if(t1.val!=t2.val)
                return false;   
            q.add(t1.left);
            q.add(t1.right);
            p.add(t2.right);
            p.add(t2.left);
        }
        return true;
        
    }
}