[LeetCode] 102. Binary Tree Level Order Traversal

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本文介绍了LeetCode上关于二叉树层次遍历的问题，并提供了两种解决方案：一种是使用单个队列并记录每层节点数量；另一种是利用两个队列进行交替存储，实现了从左到右按层次遍历二叉树节点。

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

题目

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

思路

本质是个广度遍历，两种方法。
1.为了分层会记录当前层queue的个数，每次循环只访问这些结点，将其子结点插入。
2.用两个queue，来实现交替。

code

############ 方法一 ##############

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

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """

        queue = []
        res = []
        if root!=None: 
            queue.append(root)

        while len(queue) != 0:
            queuelen = len(queue)
            tmpres  = []
            for i in range(queuelen):
                node = queue.pop(0)
                tmpres.append(node.val)
                if node.left != None:
                    queue.append(node.left)
                if  node.right != None:
                    queue.append(node.right)
            res.append(tmpres)
        return res

################# 方法二 ##############

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

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """

        queueA = []
        queueB = []
        res = []
        if root!=None: 
            queueA.append(root)

        while len(queueA) != 0 or len(queueB) != 0:
            tmpres  = []
            if len(queueA) != 0:
                while len(queueA) != 0:
                    node = queueA.pop(0)
                    tmpres.append(node.val)
                    if node.left != None:
                        queueB.append(node.left)
                    if  node.right != None:
                        queueB.append(node.right)

            else:
                while len(queueB) != 0:
                    node = queueB.pop(0)
                    tmpres.append(node.val)
                    if node.left != None:
                        queueA.append(node.left)
                    if node.right != None:
                        queueA.append(node.right)

            res.append(tmpres)
        return res