Move from brute-force thinking to an efficient approach using dynamic programming strategy.
Given an integer n, return all the structurally unique BST's (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.
Example 1:
Input: n = 3 Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]
Example 2:
Input: n = 1 Output: [[1]]
Constraints:
1 <= n <= 8Problem summary: Given an integer n, return all the structurally unique BST's (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Backtracking · Tree
3
1
unique-binary-search-trees)different-ways-to-add-parentheses)import java.util.*;
class Solution {
public List<TreeNode> generateTrees(int n) {
if (n == 0) return new ArrayList<>();
return build(1, n);
}
private List<TreeNode> build(int left, int right) {
List<TreeNode> res = new ArrayList<>();
if (left > right) {
res.add(null);
return res;
}
for (int rootVal = left; rootVal <= right; rootVal++) {
List<TreeNode> leftTrees = build(left, rootVal - 1);
List<TreeNode> rightTrees = build(rootVal + 1, right);
for (TreeNode l : leftTrees) {
for (TreeNode r : rightTrees) {
TreeNode root = new TreeNode(rootVal);
root.left = l;
root.right = r;
res.add(root);
}
}
}
return res;
}
}
func generateTrees(n int) []*TreeNode {
if n == 0 {
return []*TreeNode{}
}
return buildTrees(1, n)
}
func buildTrees(left int, right int) []*TreeNode {
if left > right {
return []*TreeNode{nil}
}
res := []*TreeNode{}
for rootVal := left; rootVal <= right; rootVal++ {
leftTrees := buildTrees(left, rootVal-1)
rightTrees := buildTrees(rootVal+1, right)
for _, l := range leftTrees {
for _, r := range rightTrees {
root := &TreeNode{Val: rootVal}
root.Left = l
root.Right = r
res = append(res, root)
}
}
}
return res
}
class Solution:
def generateTrees(self, n: int) -> List[Optional[TreeNode]]:
if n == 0:
return []
def build(left: int, right: int) -> List[Optional[TreeNode]]:
if left > right:
return [None]
res = []
for root_val in range(left, right + 1):
for l in build(left, root_val - 1):
for r in build(root_val + 1, right):
root = TreeNode(root_val)
root.left = l
root.right = r
res.append(root)
return res
return build(1, n)
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn generate_trees(n: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
if n == 0 {
return vec![];
}
fn build(left: i32, right: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
if left > right {
return vec![None];
}
let mut res = Vec::new();
for root_val in left..=right {
let left_trees = build(left, root_val - 1);
let right_trees = build(root_val + 1, right);
for l in &left_trees {
for r in &right_trees {
let node = Rc::new(RefCell::new(TreeNode::new(root_val)));
node.borrow_mut().left = l.clone();
node.borrow_mut().right = r.clone();
res.push(Some(node));
}
}
}
res
}
build(1, n)
}
}
function generateTrees(n: number): Array<TreeNode | null> {
if (n === 0) return [];
const build = (left: number, right: number): Array<TreeNode | null> => {
if (left > right) return [null];
const res: Array<TreeNode | null> = [];
for (let rootVal = left; rootVal <= right; rootVal++) {
const leftTrees = build(left, rootVal - 1);
const rightTrees = build(rootVal + 1, right);
for (const l of leftTrees) {
for (const r of rightTrees) {
const root = new TreeNode(rootVal);
root.left = l;
root.right = r;
res.push(root);
}
}
}
return res;
};
return build(1, n);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Mutable state leaks between branches.
Usually fails on: Later branches inherit selections from earlier branches.
Fix: Always revert state changes immediately after recursive call.
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.