LeetCode #107 Binary Tree Level Order Traversal II

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Problem

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

Given binary tree [3,9,20,null,null,15,7],

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  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

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

Solution1

Method: Breadth First Search

Time Complexity: O(n)

Space Complexity:

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

class Solution:
    
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []

        level = [root]
        ret = []
        
        while level:
            new_level = []
            ret.append([])
            for node in level:
                ret[-1].append(node.val)
                if node.left:
                    new_level.append(node.left)
                if node.right:
                    new_level.append(node.right)
            level = new_level
        
        return ret[::-1]

Solution2

Method: Depth First Search

Time Complexity: O(n)

Space Complexity:

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

class Solution:
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []

        stack = [(root, 0)]
        res = collections.deque([])

        while stack:
            node, level = stack.pop()
            if node:
                if len(res) < level + 1:
                    res.appendleft([])
                res[-(level+1)].append(node.val)
                if node.right:
                    stack.append((node.right, level+1))
                if node.left:
                    stack.append((node.left, level+1))
    
        return list(res)