107. Binary Tree Level Order Traversal II

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

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
function levelOrderBottom(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.unshift(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 levelOrderBottom(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 = [vals] + result
            current = nxt
        return result

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrderBottom(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([][]int{vals}, result...)
		current = next
	}

	return result
}