437. 路径总和 III

题目描述：

主要思路：

方法一：递归
从每个节点开始一次递归

class Solution {
public:
    int ans=0;
    void dfs(TreeNode* now,int targetSum, long sum)
    {
        if(!now)
            return;
        sum+=now->val;
        if(sum==targetSum)
            ans+=1;
        dfs(now->left,targetSum,sum);
        dfs(now->right,targetSum,sum);
    }
    int pathSum(TreeNode* root, int targetSum) {
        if(!root)
            return 0;
        dfs(root,targetSum,0);
        pathSum(root->left,targetSum);
        pathSum(root->right,targetSum);
        return ans;
    }
};

方法二：前缀和
利用哈希表记录前缀和，相减得到结果

class Solution {
public:
    int ans;
    unordered_map<long,int> book;
    void dfs(TreeNode* root,int targetSum, long sum)
    {
        if(!root)
            return;
        sum+=root->val;
        ans+=book[sum-targetSum];
        book[sum]+=1;
        dfs(root->left,targetSum,sum);
        dfs(root->right,targetSum,sum);
        --book[sum];
    }
    int pathSum(TreeNode* root, int targetSum) {
        book[0]=1;
        dfs(root,targetSum,0);
        return ans;
    }
};