LeetCode #1008 — MEDIUM

Construct Binary Search Tree from Preorder Traversal

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack · Tree

Example 1

[8,5,1,7,10,12]

Example 2

[1,3]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1008: Construct Binary Search Tree from Preorder Traversal
/**
 * 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[] preorder;

    public TreeNode bstFromPreorder(int[] preorder) {
        this.preorder = preorder;
        return dfs(0, preorder.length - 1);
    }

    private TreeNode dfs(int i, int j) {
        if (i > j) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[i]);
        int l = i + 1, r = j + 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (preorder[mid] > preorder[i]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        root.left = dfs(i + 1, l - 1);
        root.right = dfs(l, j);
        return root;
    }
}

// Accepted solution for LeetCode #1008: Construct Binary Search Tree from Preorder Traversal
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func bstFromPreorder(preorder []int) *TreeNode {
	var dfs func(i, j int) *TreeNode
	dfs = func(i, j int) *TreeNode {
		if i > j {
			return nil
		}
		root := &TreeNode{Val: preorder[i]}
		l, r := i+1, j+1
		for l < r {
			mid := (l + r) >> 1
			if preorder[mid] > preorder[i] {
				r = mid
			} else {
				l = mid + 1
			}
		}
		root.Left = dfs(i+1, l-1)
		root.Right = dfs(l, j)
		return root
	}
	return dfs(0, len(preorder)-1)
}

# Accepted solution for LeetCode #1008: Construct Binary Search Tree from Preorder Traversal
class Solution:
    def bstFromPreorder(self, preorder: List[int]) -> Optional[TreeNode]:
        def dfs(i: int, j: int) -> Optional[TreeNode]:
            if i > j:
                return None
            root = TreeNode(preorder[i])
            l, r = i + 1, j + 1
            while l < r:
                mid = (l + r) >> 1
                if preorder[mid] > preorder[i]:
                    r = mid
                else:
                    l = mid + 1
            root.left = dfs(i + 1, l - 1)
            root.right = dfs(l, j)
            return root

        return dfs(0, len(preorder) - 1)

// Accepted solution for LeetCode #1008: Construct Binary Search Tree from Preorder Traversal
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn bst_from_preorder(preorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        fn dfs(preorder: &Vec<i32>, i: usize, j: usize) -> Option<Rc<RefCell<TreeNode>>> {
            if i > j {
                return None;
            }
            let root = Rc::new(RefCell::new(TreeNode::new(preorder[i])));
            let mut l = i + 1;
            let mut r = j + 1;
            while l < r {
                let mid = (l + r) >> 1;
                if preorder[mid] > preorder[i] {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            let mut root_ref = root.borrow_mut();
            root_ref.left = dfs(preorder, i + 1, l - 1);
            root_ref.right = dfs(preorder, l, j);
            Some(root.clone())
        }

        dfs(&preorder, 0, preorder.len() - 1)
    }
}

// Accepted solution for LeetCode #1008: Construct Binary Search Tree from Preorder Traversal
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function bstFromPreorder(preorder: number[]): TreeNode | null {
    const dfs = (i: number, j: number): TreeNode | null => {
        if (i > j) {
            return null;
        }
        const root = new TreeNode(preorder[i]);
        let [l, r] = [i + 1, j + 1];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (preorder[mid] > preorder[i]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        root.left = dfs(i + 1, l - 1);
        root.right = dfs(l, j);
        return root;
    };
    return dfs(0, preorder.length - 1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.