作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
题目地址：https://leetcode.com/problems/path-sum-iii/#/description
题目描述
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.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
题目大意
找到二叉树中从某个顶点向下找的路径中，有多少条等于sum.
解题方法
DFS + DFS
使用DFS解决。dfs函数有两个参数，一个是当前的节点，另一个是要得到的值。当节点的值等于要得到的值的时候说明是一个可行的解。再求左右的可行的解的个数，求和之后是所有的。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int pathSum(TreeNode root, int sum) {
if(root == null){
return 0;
}
return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
public int dfs(TreeNode root, int sum){
int res = 0;
if(root == null){
return res;
}
if(root.val == sum){
res++;
}
res += dfs(root.left, sum - root.val);
res += dfs(root.right, sum - root.val);
return res;
}
}
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 = None
class Solution(object):
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: int
"""
if not root: return 0
return self.dfs(root, sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)
def dfs(self, root, sum):
res = 0
if not root: return res
sum -= root.val
if sum == 0:
res += 1
res += self.dfs(root.left, sum)
res += self.dfs(root.right, sum)
return res
BFS + DFS
使用BFS找到每个顶点作为起点的情况下，用dfs计算等于sum的路径个数。
# 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
"""
res = [0]
que = collections.deque()
que.append(root)
while que:
node = que.popleft()
if not node:
continue
self.dfs(node, res, 0, sum)
que.append(node.left)
que.append(node.right)
return res[0]
def dfs(self, root, res, path, target):
if not root: return
path += root.val
if path == target:
res[0] += 1
self.dfs(root.left, res, path, target)
self.dfs(root.right, res, path, target)
日期
2017 年 5 月 2 日
2018 年 11 月 20 日 —— 真是一个好天气