算法

https://leetcode.com/problems/path-sum-iii/description/

求和为sum的树路径数量,只能从上往下,起点不一定为根

增加一个父节点S,往下遍历时用hashmap记录S到每个点的距离,以及总距离。用总距离减去目标值即为这条路径剩余量,然后在hashmap中搜索剩余量得到路径数。

class Solution {
    public int pathSum(TreeNode root, int sum) {
        Map<Integer,Integer> map = new HashMap<>();
        map.put(0,1);
        return dfs(root,map,0,sum);
    }
    
    public int dfs(TreeNode root,Map<Integer,Integer> map,int distance,int sum)
    {
        if(root==null) return 0;
        distance += root.val;
        int numOfPath = map.getOrDefault(distance - sum,0);
        map.put(distance,map.getOrDefault(distance,0)+1);
        numOfPath += dfs(root.left,map,distance,sum);
        numOfPath += dfs(root.right,map,distance,sum);
        map.put(distance,map.get(distance)-1);
        return numOfPath;
    }
    
}

其中这两句顺序是重要的,之前这里没注意就卡了很久

int numOfPath = map.getOrDefault(distance - sum,0);
map.put(distance,map.getOrDefault(distance,0)+1);
原文地址:https://www.cnblogs.com/qlky/p/7879855.html