completed trees-2

[yashhh-23]

No description provided.

[Copilot]

Pull request overview

This PR adds Java solutions for two tree problems in the Trees-2 repository (constructing a binary tree from inorder/postorder traversals, and summing root-to-leaf numbers).

Changes:

• Added a solution for constructing a binary tree from inorder + postorder traversals.
• Added a solution for summing all root-to-leaf numbers via DFS accumulation.

Reviewed changes

Copilot reviewed 2 out of 2 changed files in this pull request and generated 3 comments.

File	Description
TreeNode.java	Adds the inorder/postorder reconstruction solution (currently missing required type/imports for standalone compilation).
sumroottoleaf.java	Adds the root-to-leaf sum solution (currently duplicates the Solution class name used elsewhere).

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[super30admin]

Construct Binary Tree from Inorder and Postorder Traversal (TreeNode.java)

Strengths:

• Clean implementation that mirrors the optimal approach
• Good documentation with time/space complexity analysis
• Correct understanding of postorder traversal (right subtree before left)
• Proper use of instance variables for state management across recursive calls

Areas for Improvement:

• Consider adding null checks for edge cases (though constraints guarantee valid input)
• Could add a brief comment explaining why right subtree is built before left (key insight for this problem)
• The solution is already quite polished; minor improvements would be cosmetic

The solution demonstrates solid understanding of tree traversals and recursive problem-solving. The approach is correct, efficient, and well-documented.

VERDICT: PASS

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Sum Root to Leaf Numbers (sumroottoleaf.java)

Strengths:

• The solution correctly implements the core algorithm
• Good use of comments to explain the approach and any difficulties faced
• Clean, readable code structure
• Proper handling of edge cases (null root)

Areas for Improvement:

• Space complexity clarification: The recursion stack space is O(h) (height of tree), not O(n). In the worst case of a skewed tree, h = n, but in balanced cases, h = log(n). This is a minor point but worth understanding the distinction.
• The solution is already optimal; no significant optimizations needed.

VERDICT: PASS