437. Path Sum III（路径可以任意点开始，任意点结束 or. 前缀和）

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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 -> 1

 

 

思路是「枚举起点 + 向下累加」：

• 内层 dfs(node, cur_sum)：固定起点为 node，统计有多少条向下路径和为 targetSum
• 外层 pathSum：把每个节点都当作一次起点，再加上左右子树的递归，共 n 个起点

代价：每个起点往下最多走 O(n)，总时间 O(n²)（平衡树约 O(n log n)），空间 O(h) 递归栈。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
        
        def dfs(root,cur_sum):
            if root == None:
                return 0 
            
            cnt = 0
            if cur_sum+root.val == targetSum:
                cnt = 1
            l = dfs(root.left,cur_sum+root.val)
            r = dfs(root.right,cur_sum+root.val)
            return l + r + cnt
        
        if root == None:
            return 0
        a = dfs(root,0)
        b = self.pathSum(root.left,targetSum)
        c = self.pathSum(root.right,targetSum)
        return a+b+c

 

/**
 * Definition

 

 

https://leetcode.com/problems/path-sum-iii/discuss/141424/Python-step-by-step-walk-through.-Easy-to-understand.-Two-solutions-comparison.-%3A-)

 

 

 

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        self.res = 0
        self.tmap = {}
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
        def dfs(root,cursum,target):
            if root == None:
                return
            cursum += root.val
            sum2 = cursum - target
            self.res += self.tmap.get(sum2,0)
            self.tmap[cursum] = self.tmap.get(cursum,0) + 1
            dfs(root.left,cursum,target)
            dfs(root.right,cursum,target)
            self.tmap[cursum]-=1
        self.tmap[0] = 1
        dfs(root,0,targetSum)
        return self.res