1008. Construct Binary Search Tree from Preorder Traversal

Problem

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.

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int i = 0;

    public TreeNode bstFromPreorder(int[] preorder) {
        return bstFromPreorder(preorder, Integer.MAX_VALUE);
    }

    private TreeNode bstFromPreorder(int[] preorder, int bound) {
        if (i == preorder.length || preorder[i] > bound) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[i++]);
        root.left = bstFromPreorder(preorder, root.val);
        root.right = bstFromPreorder(preorder, bound);
        return root;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).