102. Binary Tree Level Order Traversal

https://leetcode.com/problems/binary-tree-level-order-traversal/

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
function levelOrder(root) {
  if (!root) {
    return []
  }

  const result = []
  let current = [root]

  while (current.length) {
    let vals = []
    let next = []

    current.forEach(node => {
      vals.push(node.val)
      if (node.left) {
        next.push(node.left)
      }
      if (node.right) {
        next.push(node.right)
      }
    })

    result.push(vals)
    current = next
  }

  return result
}

Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution(object):

    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        current, result = [root], []
        while current:
            vals, nxt = [], []
            for node in current:
                vals.append(node.val)
                if node.left:
                    nxt.append(node.left)
                if node.right:
                    nxt.append(node.right)
            result.append(vals)
            current = nxt
        return result

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) [][]int {
    if root == nil {
    	return [][]int{}
    }

    result := [][]int{}
    current := []*TreeNode{root}

    for len(current) > 0 {
    	var vals []int
    	var next []*TreeNode

    	for _, node := range current {
    		vals = append(vals, node.Val)
    		if node.Left != nil {
    			next = append(next, node.Left)
    		}
    		if node.Right != nil {
    			next = append(next, node.Right)
    		}
    	}

    	result = append(result, vals)
    	current = next
    }

    return result
}