LeetCode-Binary Tree Level Order Traversal

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disappearedgod

最新推荐文章于 2024-02-06 03:28:26 发布

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本文链接：https://blog.youkuaiyun.com/disappearedgod/article/details/38897733

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本文提供了一种使用两个队列解决二叉树层次遍历的方法，并详细解释了如何实现该算法。通过实例展示了如何将二叉树的节点按层次顺序输出，适合对数据结构和算法感兴趣的读者。

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作者：disappearedgod

文章出处：http://blog.youkuaiyun.com/disappearedgod/article/details/38897733

时间：2014-8-28

题目

Binary Tree Level Order Traversal

Total Accepted: 19795 Total Submissions: 64180 My Submissions

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

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

解法

解法很简单：用两个队列，一个解决list添加问题，一个解决把list添加进去。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        ArrayList<ArrayList<Integer>> retlist = new ArrayList<ArrayList<Integer>>();
        if(root == null){
            //retlist.add(list);
            return retlist;
        }
        Queue<TreeNode> q =new LinkedList<TreeNode>();
        
        q.offer(root);
        while(!q.isEmpty()){
            Queue<TreeNode> next =new LinkedList<TreeNode>();
            list = new ArrayList<Integer>();
            while(!q.isEmpty()){
                TreeNode t = q.poll();
                list.add(t.val);
                if(t.left != null)
                    next.offer(t.left);
                if(t.right != null)
                    next.offer(t.right);
            }
            q = next;
            retlist.add(list);
        }
        return retlist;
    }
}