How do you solve Merge BSTs to Create Single BST on LeetCode?

Merge BSTs to Create Single BST (LeetCode #1932) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Merging requires careful value checking to maintain BST properties.

What is the time complexity of Merge BSTs to Create Single BST?

The optimal solution for Merge BSTs to Create Single BST runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse each tree once to collect values and then merge efficiently based on those values.

What is the brute force solution for Merge BSTs to Create Single BST?

The brute force approach for Merge BSTs to Create Single BST is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try merging each pair of trees repeatedly until only one tree remains. This approach checks every possible merge, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Merge BSTs to Create Single BST?

Clarify the merging conditions upfront. Discuss edge cases, like trees with only one node. Practice tree traversal techniques to enhance understanding.

What are common mistakes when solving Merge BSTs to Create Single BST?

Not checking for valid merges before attempting. Overlooking the BST properties during merges.

#1932

Merge BSTs to Create Single BST

Hard

ArrayHash TableTreeDepth-First SearchBinary Search TreeBinary TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking all pairs, we can build a single tree by maintaining a set of values and merging directly when possible. This reduces unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Create a set to store all values from the trees.
2Step 2: For each tree, check if its leaves can be merged with the root of another tree using the set.
3Step 3: If a merge is possible, perform it and update the set accordingly.
4Step 4: Continue until only one tree remains or no more merges can be performed.

solution.py31 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def merge_trees(trees):
9    values = set()
10    for tree in trees:
11        collect_values(tree, values)
12    while len(trees) > 1:
13        merged = False
14        for i in range(len(trees)):
15            for j in range(len(trees)):
16                if i != j and trees[j].val in values:
17                    trees[i] = merge(trees[i], trees[j])
18                    trees.pop(j)
19                    merged = True
20                    break
21            if merged:
22                break
23    return trees[0] if trees else None
24
25def collect_values(node, values):
26    if node:
27        values.add(node.val)
28        collect_values(node.left, values)
29        collect_values(node.right, values)
30
31# Helper functions can_merge and merge would be defined here.

ℹ

Complexity note: The time complexity is O(n) because we only traverse each tree once to collect values and then merge efficiently based on those values.

1Merging requires careful value checking to maintain BST properties.
2Understanding tree structures is crucial for efficient merging.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.