Leetcode: Binary Tree Level Order Traversal

Question

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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

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Analysis

using queue

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Solution

Solution 1

using two list to store node in last and current level.fa

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

class Solution:
    # @param {TreeNode} root
    # @return {integer[][]}
    def levelOrder(self, root):
        result = []
        if root==None:
            return result

        Node = [root]
        while Node!=[]:
            lastLevel, curNode = [],[]
            for temp in Node:
                lastLevel.append(temp.val)
                if temp.left!=None:
                    curNode.append(temp.left)
                if temp.right!=None:
                    curNode.append(temp.right)

            result.append(lastLevel)
            Node = curNode

        return result

Solution 2

Unlike solution 1, it uses one list and two flag to record the position in order to save storage.

class Solution:
    # @param {TreeNode} root
    # @return {integer[][]}
    def levelOrder(self, root):
        result = []
        if root==None:
            return result

        queue = [root]
        lastnum, curnum = 1, 0
        while queue!=[]:
            curresult, curnum = [], 0
            while lastnum>0:
                curresult.append(queue[0].val)
                lastnum -= 1
                if queue[0].left!=None:
                    queue.append(queue[0].left)
                    curnum  += 1
                if queue[0].right!=None:            
                    queue.append(queue[0].right)
                    curnum += 1
                queue = queue[1:]

            lastnum = curnum
            result.append(curresult)

        return result