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LeetCode 102. Binary Tree Level Order Traversal
原题
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]]
解题思路
思路一
- 先递归求出该树的深度，接着根据深度初始化结果的数组，最后通过递归，将每一层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# 递归方法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.ret def helper(self, node, dep): if node == None: return self.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 0 return max(self.depth(node.left), self.depth(node.right)) + 1
思路二
- 利用栈来实现，将每一层的节点压入栈中，然后通过迭代遍历出每一层节点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，栈 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 = http://www.mamicode.com/[]>
LeetCode 102. Binary Tree Level Order Traversal
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