给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

解题思路：

链接：https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/solutions/107105/dong-hua-yan-shi-105-cong-qian-xu-yu-zhong-xu-bian/
来源：力扣（LeetCode）
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代码：

/*** 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 {
public:TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {if (preorder.empty() || inorder.empty()) return nullptr;TreeNode * subroot = new TreeNode(preorder[0]);for(int i = 0; i < inorder.size() ; i++){if(inorder[i]==preorder[0]){vector<int> _pre(preorder.begin()+1,preorder.begin()+i+1);vector<int> pre_(preorder.begin()+i+1,preorder.end());vector<int> _in(inorder.begin(),inorder.begin()+i);vector<int> in_(inorder.begin()+i+1,inorder.end());subroot->left = buildTree(_pre,_in);subroot->right = buildTree(pre_,in_);break;}}return subroot;}
};