路径总和Ⅲ

这题和和《为K的数组》思路一致，也是用前缀表。
代码调试过，所以还加一部分用前序遍历数组和中序遍历数组构造二叉树的代码。

#include<vector>
#include<unordered_map>
#include<iostream>
using namespace std;
//Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
private:
    unordered_map<long long, int>map;
    int dfs(TreeNode* root, long long cur, int targetSum)
    {
        if (root == NULL)
        {
            return 0;
        }
        int count = 0;
        cur += root->val;
        if (map.find(cur - targetSum) != map.end())
        {
            count += map[cur - targetSum];
        }
        map[cur]++;
        int leftcount = dfs(root->left, cur, targetSum);
        int rightcount = dfs(root->right, cur, targetSum);
        map[cur]--;//因为路径总和只是针对同一个头结点，所以不是同一个头结点时需要回溯
        return count + leftcount + rightcount;
    }
public:
    int pathSum(TreeNode* root, int targetSum) {
        map[0] = 1;
        return dfs(root, 0, targetSum);
    }
};

class tree {
private:
    TreeNode* build(vector<int>& preorder, vector<int>& inorder)
    {
        if (preorder.size() == 0)
            return NULL;
        //找到根节点
        int rootvalue = preorder[0];
        TreeNode* root = new TreeNode(rootvalue);
        //叶子节点
        if (preorder.size() == 1)
            return root;
        //区分左右子树位置
        int index = 0;
        for (int i = 0; i < inorder.size(); i++)
        {
            if (inorder[i] == rootvalue)
            {
                index = i;
                break;
            }
        }
        vector<int>left_in(inorder.begin(), inorder.begin() + index);
        vector<int>right_in(inorder.begin() + index + 1, inorder.end());
        vector<int>left_pre(preorder.begin() + 1, preorder.begin() + 1 + left_in.size());
        vector<int>right_pre(preorder.begin() + 1 + left_in.size(), preorder.end());
        root->left = build(left_pre, left_in);
        root->right = build(right_pre, right_in);
        return root;
    }
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return build(preorder, inorder);
    }
};

int main()
{
    vector<int>inorder = {3,3,-2,5,2,1,10,-3,11};
    vector<int>preorder = { 10,5,3,3,-2,2,1,-3,11 };
    int targetsum = 8;
    tree mytree;
    TreeNode* root = mytree.buildTree(preorder,inorder);
    Solution solution;
    int result = solution.pathSum(root, targetsum);
    cout << result << endl;
}