How do you solve Smallest String Starting From Leaf on LeetCode?

Smallest String Starting From Leaf (LeetCode #988) 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 lexicographical order is crucial; always compare strings at leaf nodes.

What is the time complexity of Smallest String Starting From Leaf?

The optimal solution for Smallest String Starting From Leaf runs in O(n) time and uses O(n) space. The time complexity is O(n) because we visit each node once, and the space complexity is O(n) due to the recursion stack.

What is the brute force solution for Smallest String Starting From Leaf?

The brute force approach for Smallest String Starting From Leaf is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all paths from the root to each leaf node, constructing strings as we go. At the end, we can compare all the strings to find the smallest one lexicographically. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Smallest String Starting From Leaf?

Think about how to represent paths and how to compare them. Practice DFS and understand how recursion works in tree problems. Be clear about base cases and when to update your result.

What are common mistakes when solving Smallest String Starting From Leaf?

Not handling the case where the tree has only one node. Forgetting to compare strings correctly when reaching a leaf.

#988

Smallest String Starting From Leaf

Medium

StringBacktrackingTreeDepth-First SearchBinary TreeDepth-First SearchBacktracking

LeetCode ↗

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)

We can still use DFS, but we will maintain the path in reverse order and use a stack to build the strings as we backtrack. This way, we can directly compare strings without needing to construct them multiple times.

⚙️

Algorithm

3 steps

1Step 1: Use a stack to perform a depth-first search (DFS) on the tree.
2Step 2: Maintain a string that builds up as we traverse down the tree.
3Step 3: When a leaf is reached, compare the current string with the smallest string found so far and update if necessary.

solution.py20 lines

1class TreeNode:
2    def __init__(self, val=0, left=None, right=None):
3        self.val = val
4        self.left = left
5        self.right = right
6
7class Solution:
8    def smallestFromLeaf(self, root: TreeNode) -> str:
9        self.min_string = None
10        def dfs(node, path):
11            if not node:
12                return
13            path = chr(node.val + 97) + path
14            if not node.left and not node.right:
15                if self.min_string is None or path < self.min_string:
16                    self.min_string = path
17            dfs(node.left, path)
18            dfs(node.right, path)
19        dfs(root, '')
20        return self.min_string

ℹ

Complexity note: The time complexity is O(n) because we visit each node once, and the space complexity is O(n) due to the recursion stack.

1The lexicographical order is crucial; always compare strings at leaf nodes.
2Using DFS allows us to explore all paths efficiently.

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