Leetcode #889: Construct Binary Tree from Preorder and Postorder Traversal
In this guide, we solve Leetcode #889 Construct Binary Tree from Preorder and Postorder Traversal in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given two integer arrays, preorder and postorder where preorder is the preorder traversal of a binary tree of distinct values and postorder is the postorder traversal of the same tree, reconstruct and return the binary tree. If there exist multiple answers, you can return any of them.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Array, Hash Table, Divide and Conquer, Binary Tree
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]
Python Solution
# 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 constructFromPrePost(
self, preorder: List[int], postorder: List[int]
) -> Optional[TreeNode]:
def dfs(a: int, b: int, c: int, d: int) -> Optional[TreeNode]:
if a > b:
return None
root = TreeNode(preorder[a])
if a == b:
return root
i = pos[preorder[a + 1]]
m = i - c + 1
root.left = dfs(a + 1, a + m, c, i)
root.right = dfs(a + m + 1, b, i + 1, d - 1)
return root
pos = {x: i for i, x in enumerate(postorder)}
return dfs(0, len(preorder) - 1, 0, len(postorder) - 1)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.