leetcode-107. 二叉树的层次遍历 II

题目

给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

例如：
给定二叉树 [3,9,20,null,null,15,7],

3
/ \
9 20
/ \
15 7
返回其自底向上的层次遍历为：

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

python实现

1

# 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 root == None:
            return []
        q,result = [root],[]
        while q != []:
            temp = []
            for _ in range(len(q)):
                node = q.pop(0)
                temp.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            result.insert(0,temp)
        return result

2

# 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 root == None:
            return []
        parentSize = 1
        childSize = 0
        q = [root]
        result = []
        while q != []:
            temp = []
            while parentSize>0:
                node = q.pop(0)
                temp.append(node.val)
                parentSize -= 1
                if node.left:
                    q.append(node.left)
                    childSize += 1
                if node.right:
                    q.append(node.right)
                    childSize += 1
            parentSize = childSize
            childSize = 0
            result.insert(0,temp)
        return result