lc437. 路径总和 III

两种解法

解法一：直接两个递归，但是重复的计算过多

解法二：前缀和求解！

package com.codeking.lc;

import com.codeking.Node.TreeNode;

import java.util.HashMap;
import java.util.Map;

/**
 * @author xiongjl
 * @since 2023/10/28  14:57
 */
public class lc437 {

    // 前缀和解法
    public int pathSum(TreeNode root, int targetSum) {
        // 考虑根节点
        map.put(0L, 1);
        DFS(root, targetSum, 0);
        return cnt;
    }

    Map map = new HashMap<>();
    int cnt = 0;

    void DFS(TreeNode node, long targetSum, long preSum) {
        if (node == null) {
            return;
        }
        preSum += node.val;
        cnt += map.getOrDefault(preSum-targetSum, 0);
        // 前缀和
        map.put(preSum, map.getOrDefault(preSum, 0) + 1);
        DFS(node.left, targetSum, preSum);
        DFS(node.right, targetSum, preSum);
        // 回溯
        map.put(preSum, map.get(preSum)- 1);
    }

   /* public int pathSum(TreeNode root, int targetSum) {
        if (root == null) {
            return 0;
        }
        int ret = DFS(root, targetSum);
        ret += pathSum(root.left, targetSum);
        ret += pathSum(root.right, targetSum);
        return ret;
    }*/

    /*int DFS(TreeNode node, long targetSum) {
        int cnt = 0;
        if (node == null) {
            return 0;
        }
        int val = node.val;
        if (val == targetSum) {
            cnt++;
        }
        cnt += DFS(node.left, targetSum - val);
        cnt += DFS(node.right, targetSum - val);
        return cnt;
    }*/
}