Binary tree level order traversal II

Time: O(N); Space: O(N); easy

Given a binary tree, return the bottom-up level order traversal of its nodes’ values.

(ie, from left to right, level by level from leaf to root).

Example 1:

Input: {3,9,20,#,#,15,7}

  3
 /  \
9    20
    /  \
   15   7

Output (bottom-up level order traversal):

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

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution1(object):
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if root is None:
            return []

        result, current = [], [root]
        while current:
            next_level, vals = [], []
            for node in current:
                vals.append(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            current = next_level
            result.append(vals)

        return result[::-1]

if __name__ == "__main__":
    root = TreeNode(3)
    root.left = TreeNode(9)
    root.right = TreeNode(20)
    root.right.left = TreeNode(15)
    root.right.right = TreeNode(7)
    assert Solution1().levelOrderBottom(root) == [[15, 7], [9, 20], [3]]

See also:

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