LeetCode:437路径总和Ⅲ

/**
 * 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 {

    int count = 0;
    Map<Long,Integer> prefixMap = new HashMap<>();

    public int pathSum(TreeNode root, int targetSum) {
        prefixMap.put(0L,1);
        drive(root, 0L, targetSum);
        return count;
    }
    private void drive(TreeNode node, long currSum, int targetSum){
        if(node == null){
            return;
        }
        //1.往下遍历，将当前节点加入到map集合中
        currSum += node.val;
        //2.找之前节点中是否有我们需要的值
        long oldSum = currSum - targetSum;
        if(prefixMap.containsKey(oldSum)){
            count += prefixMap.get(oldSum);
        }
        //3.将新的值存进map中
        prefixMap.put(currSum,prefixMap.getOrDefault(currSum,0) + 1);

        drive(node.left,currSum,targetSum);
        drive(node.right,currSum,targetSum);
        //4。回溯
        prefixMap.put(currSum,prefixMap.get(currSum) - 1);
    }
}

prefixMap.put(currSum, prefixMap.getOrDefault(currSum, 0) + 1);

查找currSum出现过几次，查到了就把次数拿出来；没查到，就当作0次。然后在现有基础上+1，然后重新写入map集合中。

不能使用int count = prefixMap.get(18);因为如果为null执行+1操作会报空指针异常。