Problem Statement

Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.

It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.

A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.

Example 1:

Input: preorder = [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Example 2:

Input: preorder = [1,3]
Output: [1,null,3]

Constraints:

• 1 <= preorder.length <= 100
• 1 <= preorder[i] <= 108
• All the values of preorder are unique.

Problem Link

Construct Binary Search Tree from Preorder Traversal - LeetCode

Code

/**
 * 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:
    int i=0;
    TreeNode* bstFromPreorder(vector<int>& preorder) {
        
        return build(preorder,INT_MIN,INT_MAX,preorder.size());
    }
    TreeNode* build(vector<int> &preorder,int min,int max,int size)
    {
        if(i>=size||preorder[i]>max||preorder[i]<min)
            return NULL;
        
        TreeNode *root = new TreeNode(preorder[i++]);
        root->left = build(preorder,min,root->val,size);
        root->right = build(preorder,root->val,max,size);
        
        return root;
    }
};