LeetCode #1932 — HARD

Merge BSTs to Create Single BST

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given n BST (binary search tree) root nodes for n separate BSTs stored in an array trees (0-indexed). Each BST in trees has at most 3 nodes, and no two roots have the same value. In one operation, you can:

  • Select two distinct indices i and j such that the value stored at one of the leaves of trees[i] is equal to the root value of trees[j].
  • Replace the leaf node in trees[i] with trees[j].
  • Remove trees[j] from trees.

Return the root of the resulting BST if it is possible to form a valid BST after performing n - 1 operations, or null if it is impossible to create a valid BST.

A BST (binary search tree) is a binary tree where each node satisfies the following property:

  • Every node in the node's left subtree has a value strictly less than the node's value.
  • Every node in the node's right subtree has a value strictly greater than the node's value.

A leaf is a node that has no children.

Example 1:

Input: trees = [[2,1],[3,2,5],[5,4]]
Output: [3,2,5,1,null,4]
Explanation:
In the first operation, pick i=1 and j=0, and merge trees[0] into trees[1].
Delete trees[0], so trees = [[3,2,5,1],[5,4]].

In the second operation, pick i=0 and j=1, and merge trees[1] into trees[0].
Delete trees[1], so trees = [[3,2,5,1,null,4]].

The resulting tree, shown above, is a valid BST, so return its root.

Example 2:

Input: trees = [[5,3,8],[3,2,6]]
Output: []
Explanation:
Pick i=0 and j=1 and merge trees[1] into trees[0].
Delete trees[1], so trees = [[5,3,8,2,6]].

The resulting tree is shown above. This is the only valid operation that can be performed, but the resulting tree is not a valid BST, so return null.

Example 3:

Input: trees = [[5,4],[3]]
Output: []
Explanation: It is impossible to perform any operations.

Constraints:

  • n == trees.length
  • 1 <= n <= 5 * 104
  • The number of nodes in each tree is in the range [1, 3].
  • Each node in the input may have children but no grandchildren.
  • No two roots of trees have the same value.
  • All the trees in the input are valid BSTs.
  • 1 <= TreeNode.val <= 5 * 104.
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given n BST (binary search tree) root nodes for n separate BSTs stored in an array trees (0-indexed). Each BST in trees has at most 3 nodes, and no two roots have the same value. In one operation, you can: Select two distinct indices i and j such that the value stored at one of the leaves of trees[i] is equal to the root value of trees[j]. Replace the leaf node in trees[i] with trees[j]. Remove trees[j] from trees. Return the root of the resulting BST if it is possible to form a valid BST after performing n - 1 operations, or null if it is impossible to create a valid BST. A BST (binary search tree) is a binary tree where each node satisfies the following property: Every node in the node's left subtree has a value strictly less than the node's value. Every node in the node's right subtree has a value strictly greater than the node's value. A leaf is a node that has no children.

Baseline thinking

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

Pattern signal: Array · Hash Map · Tree

Example 1

[[2,1],[3,2,5],[5,4]]

Example 2

[[5,3,8],[3,2,6]]

Example 3

[[5,4],[3]]
Step 02

Core Insight

What unlocks the optimal approach

  • Is it possible to have multiple leaf nodes with the same values?
  • How many possible positions are there for each tree?
  • The root value of the final tree does not occur as a value in any of the leaves of the original tree.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #1932: Merge BSTs to Create Single BST
class Solution {
  public TreeNode canMerge(List<TreeNode> trees) {
    Map<Integer, TreeNode> valToNode = new HashMap<>(); // {val: node}
    Map<Integer, Integer> count = new HashMap<>();      // {val: freq}

    for (TreeNode tree : trees) {
      valToNode.put(tree.val, tree);
      count.merge(tree.val, 1, Integer::sum);
      if (tree.left != null)
        count.merge(tree.left.val, 1, Integer::sum);
      if (tree.right != null)
        count.merge(tree.right.val, 1, Integer::sum);
    }

    for (TreeNode tree : trees)
      if (count.get(tree.val) == 1) {
        if (isValidBST(tree, null, null, valToNode) && valToNode.size() <= 1)
          return tree;
        return null;
      }

    return null;
  }

  private boolean isValidBST(TreeNode tree, TreeNode minNode, TreeNode maxNode,
                             Map<Integer, TreeNode> valToNode) {
    if (tree == null)
      return true;
    if (minNode != null && tree.val <= minNode.val)
      return false;
    if (maxNode != null && tree.val >= maxNode.val)
      return false;
    if (tree.left == null && tree.right == null && valToNode.containsKey(tree.val)) {
      final int val = tree.val;
      tree.left = valToNode.get(val).left;
      tree.right = valToNode.get(val).right;
      valToNode.remove(val);
    }

    return                                                 //
        isValidBST(tree.left, minNode, tree, valToNode) && //
        isValidBST(tree.right, tree, maxNode, valToNode);
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.

Related Problems
#105Construct Binary Tree from Preorder and Inorder Traversalmedium→#106Construct Binary Tree from Inorder and Postorder Traversalmedium→#690Employee Importancemedium→#889Construct Binary Tree from Preorder and Postorder Traversalmedium→#1110Delete Nodes And Return Forestmedium→