LeetCode 102. Binary Tree Level Order Traversal

Problem

LeetCode 102. Binary Tree Level Order Traversal

1. 题目简述

给出一棵树，返回BFS（广度优先遍历）的结果。

Example :
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

2. 算法思路

没啥好说的，最经典的BFS遍历，5分钟以内应该能默写。

代码如下：

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        // 二叉树的BFS
        List<List<Integer>> res = new ArrayList<>();
        
        if (root == null) {
            return res;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            List<Integer> temp = new ArrayList<>();
            int levelSize = queue.size();
            
            while (levelSize-- > 0) {
                TreeNode node = queue.poll();
                temp.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            
            res.add(temp);
        }
        
        return res;
    }
}