How do you solve Minimum Depth of Binary Tree on LeetCode?

Minimum Depth of Binary Tree (LeetCode #111) 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: The minimum depth is defined by the first leaf node encountered.

What is the brute force solution for Minimum Depth of Binary Tree?

The brute force approach for Minimum Depth of Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all paths from the root to the leaves and counting the number of nodes along each path. We then find the minimum count among these paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Depth of Binary Tree?

Minimum Depth of Binary Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Tree Traversal, Breadth-First Search.

What are the interview tips for Minimum Depth of Binary Tree?

Clarify the definition of a leaf node with the interviewer. Discuss both DFS and BFS approaches to show versatility. Consider edge cases like an empty tree or a tree with only one node.

What are common mistakes when solving Minimum Depth of Binary Tree?

Not checking for an empty tree initially. Confusing depth with height; remember depth is the number of nodes, height is the number of edges.

#111

Minimum Depth of Binary Tree

Easy

TreeDepth-First SearchBreadth-First SearchBinary TreeTree TraversalBreadth-First Search

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a breadth-first search (BFS) to explore the tree level by level. The first time we encounter a leaf node, we can return its depth as the minimum depth.

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Algorithm

4 steps

1Step 1: If the tree is empty, return 0.
2Step 2: Initialize a queue and add the root node along with its depth (1).
3Step 3: While the queue is not empty, dequeue a node and check if it's a leaf. If it is, return its depth.
4Step 4: If not a leaf, enqueue its children with incremented depth.

solution.py14 lines

1from collections import deque
2
3def minDepth(root):
4    if not root:
5        return 0
6    queue = deque([(root, 1)])
7    while queue:
8        node, depth = queue.popleft()
9        if not node.left and not node.right:
10            return depth
11        if node.left:
12            queue.append((node.left, depth + 1))
13        if node.right:
14            queue.append((node.right, depth + 1))

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Complexity note: The complexity is O(n) because we may need to visit every node in the tree. The space complexity is O(n) due to the queue storing nodes at each level.

1The minimum depth is defined by the first leaf node encountered.
2BFS is effective for level-order traversal, making it suitable for this problem.

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