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
遍历整棵树，每个节点找一遍
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { int cnt = 0; public int pathSum(TreeNode root, int sum) { if (root == null) return 0; find(root, sum); pathSum(root.left, sum); pathSum(root.right, sum); return cnt; } private void find(TreeNode node, int sum) { if (node == null) return; if (sum == node.val) cnt ++; find(node.left, sum-node.val); find(node.right, sum-node.val); } }
原文：https://www.cnblogs.com/wxisme/p/9792134.html