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
题目含义：给定一棵二叉树，以及一个和sum，问在树中是否存在一条路径，路径上所有结点之和恰好等于sum。路径不限制一定要开始于根结点或者结束于叶子结点，但是一定要是向下的（从父亲结点向儿子结点）。如果存在这样的路径，求出有多少条
1 int findPath(TreeNode node, int curSum, int sum) { 2 if (node == null) return 0; 3 curSum += node.val; 4 int sameCount = curSum == sum ? 1 : 0; 5 return sameCount + findPath(node.left, curSum, sum) + findPath(node.right, curSum, sum); 6 } 7 8 public int pathSum(TreeNode root, int sum) { 9 // 以每一个节点作为路径根节点进行前序遍历，查找每一条路径的权值和与sum是否相等 10 if (root == null) return 0; 11 int res = findPath(root, 0, sum) + pathSum(root.left, sum) + pathSum(root.right, sum); 12 return res; 13 }