力扣-105题 从前序与中序遍历序列构造二叉树（C++）- dfs

题目链接：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
题目如下：

/**
 * 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) {
        //前序(根左右)  中序(左根右)
        if(preorder.size()==0||inorder.size()==0) return nullptr;
        //6个参数分别为，前序-前开始-前末尾，中序-中开始-中末尾
        return dfs(preorder,0,preorder.size(),inorder,0,inorder.size());
    }
    TreeNode* dfs(vector<int>& pre,int preBegin,int preEnd,vector<int>& in,int inBegin,int inEnd){
        //1、先序
        if(preBegin==preEnd) return nullptr;//空节点

        int rootval=pre[preBegin];
        TreeNode* root=new TreeNode(rootval);

        if(preEnd-preBegin==1) return root;//只有一个节点

        //2、中序
        int index;
        for(index=inBegin;index<inEnd;index++){
            if(in[index]==rootval) break;//找到先序节点对应在中序中所在的位置
        }

        //3、分割
        //前序
        int leftpreBegin=preBegin+1;
        int leftpreEnd=preBegin+1+index-inBegin;
        int rightpreBegin=preBegin+1+(index-inBegin);
        int rightpreEnd=preEnd;
        //中序
        int leftinBegin=inBegin;
        int leftinEnd=index;
        int rightinBegin=index+1;
        int rightinEnd=inEnd;

        root->left=dfs(pre,leftpreBegin,leftpreEnd,in,leftinBegin,leftinEnd);
        root->right=dfs(pre,rightpreBegin,rightpreEnd,in,rightinBegin,rightinEnd);

        return root;
    }
};