Construct Binary Tree from Preorder and Inorder Traversal

트리를 preorder 순서와, inorder 순서로 순회한 결과 벡터들을 받고,

해당 데이터들로 트리 구조를 파악하여 트리를 만드는 문제이다.

preorder는 root가 가장 먼저(pre) => root, left, right

inorder는 root가 중간(in) => left, root, right

/**
 * 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* DFS(vector<int>& preorder, vector<int>& inorder)
    {
        TreeNode* root = new TreeNode(preorder.front());

        if (preorder.size() == 1)
        {
            return root;
        }
        
        auto rootIt = find(inorder.begin(), inorder.end(), root->val);
        int rootIndex = distance(inorder.begin(), rootIt);
        
        vector<int> leftTreePreorder(preorder.begin() + 1, preorder.begin() + rootIndex + 1);
        vector<int> rightTreePreorder(preorder.begin() + rootIndex + 1, preorder.end());

        vector<int> leftTreeInorder(inorder.begin(), inorder.begin() + rootIndex);
        vector<int> rightTreeInorder(inorder.begin() + rootIndex + 1, inorder.end());

        if (leftTreeInorder.empty() == false)
        {
            root->left = DFS(leftTreePreorder, leftTreeInorder);
        }
        
        if (rightTreeInorder.empty() == false)
        {
            root->right = DFS(rightTreePreorder, rightTreeInorder);
        }
        
        return root;
    }
    
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {

        TreeNode *root = DFS(preorder, inorder);
        
        return root;
    }
};