给定一棵二叉树，其中每个节点都含有一个整数数值(该值或正或负)。设计一个算法，打印节点数值总和等于某个给定值的所有路径的数量。注意，路径不一定非得从二叉树的根节点或叶节点开始或结束，但是其方向必须向下(只能从父节点指向子节点方向)。示例:
给定如下二叉树，以及目标和 sum = 22，5/ \4   8/   / \11  13  4/  \    / \7    2  5   1
返回:3
解释：和为 22 的路径有：[5,4,11,2], [5,8,4,5], [4,11,7]

代码

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {int tar, cnt = 0;public int pathSum(TreeNode root, int sum) {tar = sum;Sum(root, 0, new LinkedList<>());return cnt;}public void Sum(TreeNode root, int sum, LinkedList<Integer> res) {//dfsif (root == null) return;res.add(root.val);cnt += um(res);Sum(root.left, sum + root.val, res);Sum(root.right, sum + root.val, res);res.removeLast();}public int um(LinkedList<Integer> res) {//找当前数组符合的连续子数组int temp = 0, r = 0;for (int i = res.size() - 1; i >= 0; i--) {r += res.get(i);if (r == tar) temp++;}return temp;}
}

递归+前缀和+回溯代码

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {int tar;public int pathSum(TreeNode root, int sum) {tar = sum;HashMap<Integer,Integer> map=new HashMap<>();map.put(0,1);//节点本身也算路径return helper(root, 0,map);}public int helper(TreeNode root, int sum,  HashMap<Integer,Integer> map) {if (root == null) return 0;int cnt=0;int res=sum+root.val;cnt +=map.getOrDefault(res-tar,0);//符合的前缀和出现的次数就是路径的数量map.put(res,map.getOrDefault(res,0)+1);if(root.left!=null) cnt+=helper(root.left, res, map);if(root.right!=null) cnt+=helper(root.right, res, map);map.put(res,map.get(res)-1);//回溯return cnt;}
}