从前序遍历与中序遍历序列构造二叉树

前序遍历：中左右
中序遍历：左中右
前序遍历的第一个数必定为根节点，再到中序遍历中找到该数，数的左边是左子树，右边是右子树，进行递归即可。

#include<vector>
using namespace std;

 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:
    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> preorder = { 3,9,20,15,7 };
    vector<int> inorder = { 9,3,15,20,7 };

    Solution solution;
    TreeNode* root=solution.buildTree(preorder, inorder);
}