leetcode 路径总和III java

参考leetcode上大神的思路：https://leetcode.cn/problems/path-sum-iii/solutions/596361/dui-qian-zhui-he-jie-fa-de-yi-dian-jie-s-dey6，添加了自己的注释。
前缀和为Long类型 Map<Long,Integer> prefixSumCount = new HashMap<>();
recur函数里要有prefixSumCount和targetSum,或者宏观定义一下。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution { public int pathSum(TreeNode root, int targetSum) {Map<Long,Integer> prefixSumCount = new HashMap<>();prefixSumCount.put(0L,1);//prefixMap: key--前缀和 value--前缀和的节点数量return recur(root,prefixSumCount,targetSum,0L);}private int recur(TreeNode node, Map<Long, Integer> prefixSumCount, int targetSum, Long curSum){if(node == null){return 0;}int res = 0;curSum += node.val;//看看之前的前缀和有没有为curSum-targetSum的res += prefixSumCount.getOrDefault(curSum-targetSum,0);//更新key为curSum的value值prefixSumCount.put(curSum, prefixSumCount.getOrDefault(curSum,0)+1);//更新左右子节点int left = recur(node.left,prefixSumCount,targetSum,curSum);int right = recur(node.right,prefixSumCount,targetSum,curSum);res = res+left+right;//恢复状态： 在遍历完一个节点的所有子节点后，将其从map中除去。prefixSumCount.put(curSum,prefixSumCount.get(curSum)-1);return res;}}