问题描述
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
给定一个二叉树,其中每个节点都包含一个整数值。
找出与给定值相加的路径数。
路径不需要在根节点或叶节点处开始或结束,但必须向下(只从父节点移动到子节点)。
树的节点不超过1,000个,值的范围是-1,000,000到1,000,000。
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
返回 3.
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
Python 实现
题目要求的是路径必须从上到下,从祖先到后代,因此可以利用 BFS (深度优先搜索)的思路来遍历二叉树。
代码中定义了两个方法:pathSum 和 pathFrom。前者是从 sum 开始,以当前节点作为根节点出发遍历子树;后者是从剩余的数值 val 开始,以当前节点作为根节点出发遍历子树。因此,通过 pathFrom 实现了 BFS,伴随着 sum 的减少;通过 pathSum 实现子树根节点递归调用,以 sum 值在每个节点启动实现 BFS。从而考虑到题目要求的各种情况。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: int
"""
# DFS
if root == None:
return 0
return self.pathFrom(root, sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)
def pathFrom(self, node, val):
if node == None:
return 0
cnt = self.pathFrom(node.left, val - node.val) + self.pathFrom(node.right, val - node.val)
if node.val == val:
cnt += 1
return cnt