leetcode.107.Binary Tree Level Order Traversal II
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Description
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3]]
sln1
看到题目很直接就想到用广度优先搜索,用递归实现。python代码如下
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution(object): def levelOrderBottom(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ if root == None: return [] return self.bfs([root]) def bfs(self, nodes): if len(nodes) == 0: return [] branch = [] values = [] for node in nodes: values.append(node.val) if node.left != None: branch.append(node.left) if node.right != None: branch.append(node.right) next_layer_values = self.bfs(branch) next_layer_values.append(values) return next_layer_values
每一次递归一个循环,分别记录values和branch数组。values数组记录当前层每个节点的val,branch数组记录当前层的所有子节点。接下来调用bfs方法获取到当前层以下的所有层的遍历结果,再把values数组添加进去并返回。思路十分清晰,但是提交后看结果并不是十分理想,同样是python实现,只超过了10%+的提交。
sln2
参考discussion中一个标题为 C++ 4ms solution 的答案,用python简答复现了一次,提交结果尽然比70%+的结果快。python实现如下:
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution(object): def levelOrderBottom(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ if root == None: return [] depth = self.depth(root) return self.levelOrder([[] for _ in xrange(depth)], root, depth - 1) def levelOrder(self, ans, node, depth): if node is None: return ans ans[depth].append(node.val) ans = self.levelOrder(ans, node.left, depth - 1) ans = self.levelOrder(ans, node.right, depth - 1) return ans def depth(self, node): if node is None: return 0 return max(self.depth(node.left), self.depth(node.right)) + 1
一开始并不是很理解为什么这个实现会比我的第一种实现更快,思考一下之后发现,第一种实现方法里面很显然是有一个循环的。在这个循环中,我们遍历了一遍父节点,并把子节点都压进一个数组中,相当于也遍历了一遍子节点。当我们把这个子节点数组丢进下一次递归时,又会再遍历一次,所以实际上等于所有子节点都遍历了两次,所以速度上就慢了很多
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