How do you solve Recover a Tree From Preorder Traversal on LeetCode?

Recover a Tree From Preorder Traversal (LeetCode #1028) 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: Understanding the relationship between depth and node placement is crucial for tree reconstruction.

What is the time complexity of Recover a Tree From Preorder Traversal?

The optimal solution for Recover a Tree From Preorder Traversal runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the string once, and the space complexity is O(n) due to the stack storing nodes.

What is the brute force solution for Recover a Tree From Preorder Traversal?

The brute force approach for Recover a Tree From Preorder Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves parsing the string character by character, identifying the depth of each node based on the number of dashes, and then reconstructing the tree by maintaining a list of nodes at each depth. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Recover a Tree From Preorder Traversal?

Recover a Tree From Preorder Traversal uses the following concepts: String, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Stack, Depth-First Search.

What are the interview tips for Recover a Tree From Preorder Traversal?

Always clarify the input format and constraints before diving into the solution. Think about edge cases, such as trees with only one child or very deep trees. Practice explaining your thought process as you code; it helps solidify your understanding.

What are common mistakes when solving Recover a Tree From Preorder Traversal?

Failing to correctly track the depth of nodes, leading to incorrect parent-child assignments. Not handling cases where nodes have only one child properly.

#1028

Recover a Tree From Preorder Traversal

Hard

StringTreeDepth-First SearchBinary TreeStackDepth-First Search

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)

The optimal solution uses a stack to track nodes and their depths efficiently, allowing us to reconstruct the tree in a single pass through the string. This method is much faster and avoids redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack to hold nodes.
2Step 2: Loop through the string, counting dashes to determine the depth of each node and extracting the node value.
3Step 3: For each node, use the stack to find the correct parent based on depth, linking the node appropriately.

solution.py30 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 recoverFromPreorder(self, S: str) -> TreeNode:
9        stack = []
10        i = 0
11        while i < len(S):
12            level = 0
13            while i < len(S) and S[i] == '-':
14                level += 1
15                i += 1
16            val = 0
17            while i < len(S) and S[i].isdigit():
18                val = val * 10 + int(S[i])
19                i += 1
20            node = TreeNode(val)
21            if level == len(stack):
22                if stack:
23                    stack[-1].left = node
24                stack.append(node)
25            else:
26                while level < len(stack):
27                    stack.pop()
28                stack[-1].right = node
29                stack.append(node)
30        return stack[0]

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once, and the space complexity is O(n) due to the stack storing nodes.

1Understanding the relationship between depth and node placement is crucial for tree reconstruction.
2Using a stack effectively allows us to manage parent-child relationships without excessive searching.

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