LeetCode 106: Construct Binary Tree from Inorder and Postorder Traversal — Python Solution

Solve LeetCode 106: Construct Binary Tree from Inorder and Postorder Traversal in Python with a reverse postorder partition approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

What the problem is testing

Take roots from postorder backward and build the right subtree before the left using inorder boundaries.

Algorithm

1. Take roots from postorder backward and build the right subtree before the left using inorder boundaries.
2. Maintain this invariant: The reverse postorder cursor identifies the root of the current inorder range, followed by its right subtree.
3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

LeetCode provides the list, tree, or graph node definition used by the method.

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def buildTree(self, inorder, postorder):
        position = {value: i for i, value in enumerate(inorder)}
        post_index = len(postorder) - 1
        def build(left, right):
            nonlocal post_index
            if left > right:
                return None
            value = postorder[post_index]
            post_index -= 1
            root = TreeNode(value)
            split = position[value]
            root.right = build(split + 1, right)
            root.left = build(left, split - 1)
            return root
        return build(0, len(inorder) - 1)

Why this is correct

The proof follows the maintained state: The reverse postorder cursor identifies the root of the current inorder range, followed by its right subtree. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(n) space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

Building left first would consume roots in the wrong order.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.

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