# 437. Path Sum III 路径总和 III

@TOC

## # 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
``````

## # 解题方法

### # DFS + DFS

``````/**
* 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

``````# 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