Build Tree from Inorder and Postorder

1. Step 1: Initialize Solution.buildTree

   Replace the empty starter with the first real state owned by Solution.buildTree. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Assemble the primary transition

   Extend the initialized state with the next contiguous part of the popular solution. The transition explains how one input element or operation changes the state; boundaries are easier to reason about after this invariant is visible.

3. Step 3: Pass the Right Chain case

   Complete the readable core algorithm for one representative Interview case. Consume postorder from the end and build right before left.

4. Step 4: Harden the Sample boundary

   Repair the reviewed boundary and pass the complete submission contract. Backward postorder exposes each subtree root followed by its right then left subtree roots. Inorder positions uniquely delimit those intervals, so the recursion reconstructs the exact tree.

• Backward postorder visits root, right, left, so constructing left first consumes the wrong root.
• trees and graphs
• hashing and sets

• LeetCode 106 — Construct Binary Tree from Inorder and Postorder Traversal

• Partition Traversal Ranges(opens in a new tab)

  Partition Traversal Ranges isolates each frame describes matching node sets, and the inorder root offset determines the exact left-subtree size in preorder. That focused state discipline is required when implementing build tree inorder postorder as a complete Interview Problem.

Map inorder indices, keep a cursor at postorder end, and recursively build inclusive intervals root-right-left.

Why it works: Backward postorder exposes each subtree root followed by its right then left subtree roots. Inorder positions uniquely delimit those intervals, so the recursion reconstructs the exact tree.

class Solution:
    def buildTree(self, inorder, postorder):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        positions = {value:index for index,value in enumerate(inorder)}
        postorder_index = len(postorder) - 1
        def build(left, right):
            nonlocal postorder_index
            if left > right: return None
            value = postorder[postorder_index]
            postorder_index -= 1
            node = TreeNode(value)
            middle = positions[value]
            node.right = build(middle + 1, right)
            node.left = build(left, middle - 1)
            return node
        return build(0, len(inorder) - 1)

• Build Tree from Preorder and Inorder(opens in a new tab)