Add One Row to Tree

The problem is to add a new row of nodes at a given depth in a binary tree. Here’s a Python implementation for the addOneRow method, based on the constraints and the rules you provided:

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class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
        if depth == 1:
            return TreeNode(val, left=root)

        def helper(node, current_depth):
            if not node:
                return
            if current_depth == depth - 1:
                left_subtree = TreeNode(val, left=node.left)
                right_subtree = TreeNode(val, right=node.right)
                node.left = left_subtree
                node.right = right_subtree
            else:
                helper(node.left, current_depth + 1)
                helper(node.right, current_depth + 1)

        helper(root, 1)
        return root

This code uses a recursive helper function to traverse the tree and find the nodes at depth - 1. Then, it inserts the new nodes with the given value, attaching the original left and right subtrees to the corresponding new nodes as required. If the specified depth is 1, a new root is created, and the original tree becomes the left subtree of this new root.

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class Solution:
    def addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
        if depth == 1:
            new_root = TreeNode(val)
            new_root.left = root
            return new_root

        def dfs(node: TreeNode, current_depth: int):
            if not node:
                return
            if current_depth == depth - 1:
                left_subtree = node.left
                right_subtree = node.right
                node.left = TreeNode(val)
                node.right = TreeNode(val)
                node.left.left = left_subtree
                node.right.right = right_subtree
            else:
                dfs(node.left, current_depth + 1)
                dfs(node.right, current_depth + 1)

        dfs(root, 1)
        return root

The function adds a new row to a binary tree at a specified depth, with all nodes in that row having a specified value. Here are some components that can help to understand and categorize this problem:

This problem belongs to the domain of binary trees and tree manipulation.

What Components:

• Input:
  • root, representing the root node of a given binary tree.
  • val, an integer representing the value of the new nodes.
  • depth, an integer representing the depth at which the new row should be added.
• Output: The root node of the modified tree after adding the new row.
• Constraints: Constraints on the values of the nodes, the value to be added, and the depth of the tree.

Classifying the Problem:

• Data Structure Used: Binary Tree.
• Operations: Insertion, Traversal.
• Algorithms and Techniques: This could be solved using a level-order traversal or a depth-first traversal to find the nodes at the required depth. From there, new nodes can be inserted according to the rules described.

Slightly Simpler Problem: If you’re looking for something slightly simpler related to this problem, you might consider:

• Adding a new node at a specific position in the tree, rather than an entire row.
• Finding the nodes at a specific depth without performing the insertions.

These simpler tasks build up the skills and understanding necessary to tackle the original problem.