C++ 使用先根與後根還原二叉樹

#include <stdio.h>
#include <vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin) {
    int length = pre.size();
    if (pre.size()==0 || vin.size() == 0)
        return nullptr;
    //Create root
    TreeNode* head = (TreeNode*)malloc(sizeof(TreeNode));
    int j = 0;

    int nextInsert = 0;

    for (j; j < length; j++)
    {
        bool contains = false;
        int jj = 0;

        for (jj; jj < vin.size(); jj++)
            if (vin[jj] == pre[j])
            {
                contains = true;
                
                nextInsert = pre[j];
                break;
            }
        if (contains) break;
    }
    vector<int> newPre;
    int i_t = 0;
    for (i_t; i_t < pre.size(); i_t++)
    {
        if (pre[i_t] != nextInsert)
            newPre.push_back(pre[i_t]);
    }


    head->val = nextInsert;

    //Create left part
    vector<int> leftsubvin;
    vector<int> rightsubvin;
    int i = 0; bool left = true;
    for (i; i < vin.size(); i++)
    {
        if (vin[i] == nextInsert)
            left = false;
        else
        {
            if (left)
            {
                leftsubvin.push_back(vin[i]);
            }
            else
            {
                rightsubvin.push_back(vin[i]);
            }
        }
    }

    head->left = reConstructBinaryTree(newPre, leftsubvin);

    head->right = reConstructBinaryTree(newPre, rightsubvin);

    return head;
}

int main()
{
    //[1,2,3,4,5,6,7],[3,2,4,1,6,5,7]
    vector<int> pre = { 1,2,3,4,5,6,7 };
    vector<int> vin = { 3,2,4,1,6,5,7 };

    TreeNode * temp = reConstructBinaryTree(pre , vin);


}