输入一棵二叉树前序遍历和中序遍历的结果，请重建该二叉树。
注意:
二叉树中每个节点的值都互不相同；
输入的前序遍历和中序遍历一定合法；
样例
给定：
前序遍历是：[3, 9, 20, 15, 7]
中序遍历是：[9, 3, 15, 20, 7]
返回：[3, 9, 20, null, null, 15, 7, null, null, null, null]
返回的二叉树如下所示：
    3
   / \
  9  20
    /  \
   15   7
原题链接：https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    map<int,int> hash;
    vector<int> preorder,inorder;
    TreeNode* buildTree(vector<int>& _preorder, vector<int>& _inorder) {
        preorder = _preorder,inorder= _inorder;
        for(int i = 0; i < inorder.size();i++) hash[inorder[i]] = i;
        return dfs(0,preorder.size()-1,0,inorder.size()-1);
    }
    TreeNode* dfs(int pl,int pr,int il, int ir)
    {
        if(pl>pr) return nullptr;
        auto root = new TreeNode(preorder[pl]);
        int k = hash[root->val];
        auto left = dfs(pl+1,pl+1+k-il-1,il,k-1);
        auto right = dfs(pl+1+k-il,pr,k+1,ir);
        root->left = left;
        root->right = right;
        return root;
    }
};

vector<int> preorder,inorder;
//return dfs(0,preorder.size()-1,0,inorder.size()-1);

for(int i = 0; i< inorder.size() ; i++) hash[inorder[i]] = i;

if(pl > pr) return nullptr;

auto root = new TreeNode(preorder[pl]);

int k = hash[root->val];

auto left = dfs(pl+1,pl+k-il,il,k-1);
auto right = dfs(pl+k-il+1,pr,k+1,ir);

root->left = left, root->right = right;