BST----Binary Tree Level Order Traversal

最新推荐文章于 2020-01-11 11:22:31 发布

zjj2015

最新推荐文章于 2020-01-11 11:22:31 发布

阅读量255

点赞数

CC 4.0 BY-SA版权

分类专栏：算法

本文链接：https://blog.youkuaiyun.com/zjj2015/article/details/50547780

算法专栏收录该内容

131 篇文章

订阅专栏

本文介绍了一种解决二叉树层次遍历问题的方法，并提供了使用队列实现的具体Java代码示例。通过此方法，可以有效地按从左到右、逐层的顺序获取二叉树节点的值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Have you met this question in a real interview? Yes
Example
Given binary tree {3,9,20,#,#,15,7},

3
/ \
9 20
/ \
15 7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
]
Challenge
Challenge 1: Using only 1 queue to implement it.

Challenge 2: Use DFS algorithm to do it.

递归算法

Java queue Api

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        // write your code here
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null) return result ;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        while(!q.isEmpty()){
            int qLen = q.size();
            ArrayList<Integer> aList = new ArrayList <Integer>();
            for(int i = 0; i<qLen; i++){
                TreeNode node = q.poll();//出队
                aList.add(node.val);
                if(node.left!= null) q.offer(node.left);
                if(node.right!=null) q.offer(node.right);//入队

            }
            result.add(aList);
        }
        return result ;



    }
}