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.
Build confidence with an intuition-first walkthrough focused on tree fundamentals.
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
Note: A leaf is a node with no children.
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: 2
Example 2:
Input: root = [2,null,3,null,4,null,5,null,6] Output: 5
Constraints:
[0, 105].-1000 <= Node.val <= 1000Problem summary: Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[3,9,20,null,null,15,7]
[2,null,3,null,4,null,5,null,6]
binary-tree-level-order-traversal)maximum-depth-of-binary-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #111: Minimum Depth of Binary Tree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null) {
return 1 + minDepth(root.right);
}
if (root.right == null) {
return 1 + minDepth(root.left);
}
return 1 + Math.min(minDepth(root.left), minDepth(root.right));
}
}
// Accepted solution for LeetCode #111: Minimum Depth of Binary Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func minDepth(root *TreeNode) int {
if root == nil {
return 0
}
if root.Left == nil {
return 1 + minDepth(root.Right)
}
if root.Right == nil {
return 1 + minDepth(root.Left)
}
return 1 + min(minDepth(root.Left), minDepth(root.Right))
}
# Accepted solution for LeetCode #111: Minimum Depth of Binary Tree
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
if root.left is None:
return 1 + self.minDepth(root.right)
if root.right is None:
return 1 + self.minDepth(root.left)
return 1 + min(self.minDepth(root.left), self.minDepth(root.right))
// Accepted solution for LeetCode #111: Minimum Depth of Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
if node.left.is_none() {
return 1 + Self::dfs(&node.right);
}
if node.right.is_none() {
return 1 + Self::dfs(&node.left);
}
1 + Self::dfs(&node.left).min(Self::dfs(&node.right))
}
pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
Self::dfs(&root)
}
}
// Accepted solution for LeetCode #111: Minimum Depth of Binary Tree
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function minDepth(root: TreeNode | null): number {
if (root == null) {
return 0;
}
const { left, right } = root;
if (left == null) {
return 1 + minDepth(right);
}
if (right == null) {
return 1 + minDepth(left);
}
return 1 + Math.min(minDepth(left), minDepth(right));
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.