How do you solve Smallest Subtree with all the Deepest Nodes on LeetCode?

Smallest Subtree with all the Deepest Nodes (LeetCode #865) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: The deepest nodes can be found using a single DFS traversal.

What is the brute force solution for Smallest Subtree with all the Deepest Nodes?

The brute force approach for Smallest Subtree with all the Deepest Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves finding all the deepest nodes first and then checking each node in the tree to see if it contains all these deepest nodes. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Smallest Subtree with all the Deepest Nodes?

Explain your thought process clearly while coding. Consider edge cases and test your solution with them. Use comments to clarify your logic, especially in complex sections.

What are common mistakes when solving Smallest Subtree with all the Deepest Nodes?

Not considering edge cases like a tree with only one node. Failing to correctly identify the lowest common ancestor.

#865

Smallest Subtree with all the Deepest Nodes

Medium

Hash TableTreeDepth-First SearchBreadth-First SearchBinary TreeDepth-First SearchLowest Common Ancestor

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(h)

The optimal solution uses a single DFS traversal to find the deepest nodes and their common ancestor. This is efficient because we only traverse the tree once.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS to find the maximum depth of the tree while keeping track of the deepest nodes.
2Step 2: During the DFS, if both left and right depths are equal, the current node is a candidate for the smallest subtree containing all deepest nodes.
3Step 3: Return the node that is found to be the lowest common ancestor of the deepest nodes.

solution.py29 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
8class Solution:
9    def subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:
10        def dfs(node):
11            if not node:
12                return 0
13            left_depth = dfs(node.left)
14            right_depth = dfs(node.right)
15            if left_depth == right_depth:
16                return left_depth + 1
17            return max(left_depth, right_depth) + 1
18
19        return self.findLCA(root)
20
21    def findLCA(self, node):
22        if not node:
23            return None
24        left = self.findLCA(node.left)
25        right = self.findLCA(node.right)
26        if left and right:
27            return node
28        return left if left else right
29

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1The deepest nodes can be found using a single DFS traversal.
2The lowest common ancestor of all deepest nodes is the smallest subtree containing them.

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