[LeetCode] Binary Tree Level Order Traversal

最新推荐文章于 2025-09-18 03:57:00 发布

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#二叉树 #遍历 #python

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本文介绍了一种二叉树的层次遍历方法，并详细解释了如何通过队列实现节点值的分层输出。该算法利用父节点和子节点数量标记来确定遍历的层级。

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},

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

这题目其实是要求按层遍历二叉树，并且分层输出。在遍历二叉树时最简单就是递归了，但简单递归无法保存层的信息，在借助队列遍历时，也只能简单的按层输出，无法确

元素在那一层，这里借鉴了别人的思想：设置两个标记量分别标记parent数和child数，再说一次，要得到按层遍历的序列不难，难的是怎么确定什么时候是分层，开始的

时候queue只有root一个元素，此时parentIndex = 1, childIndex = 0，由于它的left和right都为飞空，所以root.left 和 root.right 分别入队列,

childIndex 递增两次，即childIndex += 2。。。算了，我也说不清，还是上代码吧：

class TreeNode:
	def __init__(self, x):
		self.val = x
		self.left = None
		self.right = None

class Solution:
	# @param root, a tree node
	# @param sum, an integer
	# @return a boolean
			
	def levelOrder(self, root):
		if None == root:
			return []
		queue = [root]
		ret = []
		tmp = []
		parentIndex = 1
		childIndex = 0
		while len(queue) > 0:
			leaf = queue[0]
			tmp.append(leaf.val)
			del queue[0]
			if None != leaf.left:
				queue.append(leaf.left)
				childIndex += 1
			if None != leaf.right:
				queue.append(leaf.right)
				childIndex += 1
			parentIndex -= 1

			if 0 == parentIndex:
				ret.append(tmp)
				tmp = []
				parentIndex = childIndex
				childIndex = 0
		return ret