1008. Construct Binary Search Tree from Preorder Traversal

1008. Construct Binary Search Tree from Preorder Traversal | LeetCode Solution

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] <= 1000
All the values of preorder are unique.

/**
 * 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 *bstFromPreorder(vector<int> &v)
  {
    int n = v.size();
    TreeNode *ans = new TreeNode(v[0]);
    TreeNode *curr = ans;
    int i = 1;
    stack<TreeNode *> st;
    while (i < v.size())
    {
      if (curr->val > v[i])
      {
        curr->left = new TreeNode(v[i]);
        st.push(curr);
        curr = curr->left;
      }
      else
      {
        while (!st.empty() && st.top()->val < v[i])
        {
          curr = st.top();
          st.pop();
        }
        curr->right = new TreeNode(v[i]);
        curr = curr->right;
      }
      i++;
    }
    return ans;
  }
};