How do you solve Add One Row to Tree on LeetCode?

Add One Row to Tree (LeetCode #623) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree depth is crucial for manipulating tree structures.

What is the brute force solution for Add One Row to Tree?

The brute force approach for Add One Row to Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we traverse the tree to the specified depth and then add new nodes with the given value. This method is straightforward but inefficient, as it may require multiple traversals of the tree. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Add One Row to Tree?

Add One Row to Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Tree Traversal, Level Order Traversal.

What are the interview tips for Add One Row to Tree?

Always consider edge cases, such as adding a row at the root level. Explain your thought process clearly while coding, especially when modifying tree structures. Practice tree traversal techniques, as they are commonly used in tree-related problems.

What are common mistakes when solving Add One Row to Tree?

Not handling the case where depth is 1 correctly. Forgetting to adjust the original children of nodes when adding new nodes.

#623

Add One Row to Tree

Medium

TreeDepth-First SearchBreadth-First SearchBinary TreeTree TraversalLevel Order Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single traversal to reach the desired depth, making it more efficient. We can use a queue to manage the nodes at the current depth and add new nodes in one pass.

⚙️

Algorithm

3 steps

1Step 1: If depth is 1, create a new root node with the given value and set the original tree as its left child.
2Step 2: Use a queue to perform a level-order traversal until reaching depth - 1.
3Step 3: For each node at depth - 1, create two new nodes with the given value and adjust the original children accordingly.

solution.py27 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def addOneRow(root, val, depth):
9    if depth == 1:
10        new_root = TreeNode(val)
11        new_root.left = root
12        return new_root
13    queue = [root]
14    current_depth = 1
15    while queue:
16        if current_depth == depth - 1:
17            for node in queue:
18                new_left = TreeNode(val)
19                new_right = TreeNode(val)
20                new_left.left = node.left
21                new_right.right = node.right
22                node.left = new_left
23                node.right = new_right
24            return root
25        current_depth += 1
26        queue = [child for node in queue for child in (node.left, node.right) if child]
27    return root

ℹ

Complexity note: The time complexity is O(n) because we traverse each node in the tree once. The space complexity is O(n) due to the queue used for level-order traversal.

1Understanding tree depth is crucial for manipulating tree structures.
2Level-order traversal is an effective way to access nodes at specific depths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.