LeetCode 105: Construct Binary Tree from Preorder and Inorder Traversal — Python Solution

LeetCode 105: Construct Binary Tree from Preorder and Inorder Traversal is a Medium binary tree general problem. This Python walkthrough develops an indexed recursive partition solution, ties every code decision to a concrete invariant, and includes the regression check used before publication.

This independent guide paraphrases the task rather than reproducing LeetCode’s prompt. Use the official page for the exact statement, examples, constraints, and submission runner.

Recognizing the pattern

Tree recursion works when the return value has one precise meaning for every subtree; the parent can then combine child results locally.

For this problem specifically, take roots from preorder in order and use an inorder index map to split each subtree into left and right ranges. The invariant worth writing beside the code is: The preorder cursor always identifies the root of the current inorder range.

Step-by-step algorithm

1. Identify the input state consumed by buildTree(preorder, inorder) and initialize the data required by the indexed recursive partition pattern.
2. Take roots from preorder in order and use an inorder index map to split each subtree into left and right ranges.
3. After each update, verify the page’s central invariant: The preorder cursor always identifies the root of the current inorder range.
4. Finish only after the boundary behavior is covered: Values are unique; empty inorder ranges return no node.

Python solution

LeetCode supplies the list, tree, or graph node class referenced by this method. The downloadable test suite includes compatible local node definitions so the implementation can also run outside the judge.

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

Reading the implementation

The main entry point is buildTree(preorder, inorder). The named working state includes position, pre_index, value, root, split; those variables make the indexed recursive partition state visible instead of hiding it in incidental control flow.

The method expresses the transformation directly without a general traversal loop. Early returns stop as soon as the answer is forced, avoiding work that cannot change the result. In concrete terms, take roots from preorder in order and use an inorder index map to split each subtree into left and right ranges.

Correctness argument

Base case. Empty or terminal subproblems return a result that already satisfies the claim for that smallest state.

Inductive step. Take roots from preorder in order and use an inorder index map to split each subtree into left and right ranges. Assuming child or smaller states are correct, the current call combines only results allowed by the problem, so the preorder cursor always identifies the root of the current inorder range.

Conclusion. Every recursive call reduces the remaining state. The base cases terminate, and induction carries the invariant back to the original input, establishing the returned answer.

Complexity and trade-offs

O(n) time and O(n) space. The auxiliary-space figure excludes the returned output unless the output itself is the structure being built.

An explicit stack avoids recursion-depth limits, but it must carry the same state that recursive call frames provide automatically. That comparison is useful in an interview because it explains why the final implementation is preferable, not merely that it passes.

Regression check

The published implementation belongs to a 100-problem suite that is compiled and exercised behaviorally before deployment.

One reference assertion from that suite is shown below. It targets the normal path while the edge conditions in the next section cover the failure-prone boundaries.

built=Solution().buildTree([3,9,20,15,7],[9,3,15,20,7]); assert inorder(built)==[9,3,15,20,7]

Common mistakes and edge cases

• Problem-specific boundary: Values are unique; empty inorder ranges return no node.
• Pattern-level pitfall: Write the empty-subtree result first and keep returned subtree information separate from any global answer updated at a node.
• Invariant check: after every update, confirm that the preorder cursor always identifies the root of the current inorder range.

Interview review checklist

• Explain why indexed recursive partition matches the structure of this input.
• State the invariant in one sentence before tracing code: The preorder cursor always identifies the root of the current inorder range.
• Derive O(n) time and O(n) space from how many times each element or state is visited.
• Test the boundary explicitly: Values are unique; empty inorder ranges return no node.

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