原題

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

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return its level order traversal as:

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

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解題思路

思路一

• 先遞歸求出該樹的深度，接著根據深度初始化結果的數組，最后通過遞歸，將每一層的值依次添加到答案中

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None# 遞歸方法 class Solution(object):def levelOrder(self, root):""":type root: TreeNode:rtype: List[List[int]]"""if root == None:return []dep = self.depth(root)self.ret = [[] for i in range(dep)]self.helper(root, 0)return self.retdef helper(self, node, dep):if node == None:returnself.ret[dep].append(node.val)self.helper(node.left, dep + 1)self.helper(node.right, dep + 1)def depth(self, node):if node == None:return 0return max(self.depth(node.left), self.depth(node.right)) + 1

　　

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思路二

• 利用棧來實現，將每一層的節點壓入棧中，然后通過迭代遍歷出每一層節點中的值并加入答案中

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None# 迭代方法，棧 class Solution(object):def levelOrder(self, root):""":type root: TreeNode:rtype: List[List[int]]""" if root == None:return []stack, ret = [root], []while stack:level_value = []next_level = []for node in stack:level_value.append(node.val)if node.left:next_level.append(node.left)if node.right:next_level.append(node.right)ret.append(level_value)stack = next_levelreturn ret

　　

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轉載于:https://www.cnblogs.com/LiCheng-/p/6902352.html

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