Binary Tree Level Order Traversal II

Question

• leetcode: Binary Tree Level Order Traversal II | LeetCode OJ
• lintcode: (70) Binary Tree Level Order Traversal II

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7


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

题解

此题在普通的 BFS 基础上增加了逆序输出，简单的实现可以使用辅助栈或者最后对结果逆序。

• reference

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

class Solution:
    # @param root, a tree node
    # @return a list of lists of integers
    def preorder(self, root, level, res):
        if root:
            if len(res) < level+1: res.append([])
            res[level].append(root.val)
            self.preorder(root.left, level+1, res)
            self.preorder(root.right, level+1, res)

    def levelOrderBottom(self, root):
        res=[]
        self.preorder(root, 0, res)
        res.reverse()
        return res

• reference

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        this.val = val
        this.left, this.right = None, None
"""


class Solution:
    """
    @param root: The root of binary tree.
    @return: buttom-up level order in a list of lists of integers
    """
    def levelOrderBottom(self, root):
        # write your code here
        self.results = []
        if not root:
            return self.results
        q = [root]
        while q:
            new_q = []
            self.results.append([n.val for n in q])
            for node in q:
                if node.left:
                    new_q.append(node.left)
                if node.right:
                    new_q.append(node.right)
            q = new_q
        return list(reversed(self.results))

复杂度分析

时间复杂度为 $$O(n)$$, 使用了队列或者辅助栈作为辅助空间，空间复杂度为 $$O(n)$$.