#623

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Add One Row to Tree

The problem requires adding a row of nodes to a binary tree at a specified depth.

tree depth first search breadth first search binary tree

Problem Statement

You are given the root of a binary tree and two integers, val and depth. Your task is to add a row of nodes with value val at the given depth. The root node is considered at depth 1. You need to insert the row starting at the given depth, ensuring the parent-child relationships are maintained.

If the given depth is 1, a new root node is added, and the existing root becomes the left child. If the depth is greater than 1, insert the new row between existing nodes while preserving the structure of the tree.

Examples

Example 1

Input: root = [4,2,6,3,1,5], val = 1, depth = 2

Output: [4,1,1,2,null,null,6,3,1,5]

Example details omitted.

Example 2

Input: root = [4,2,null,3,1], val = 1, depth = 3

Output: [4,2,null,1,1,3,null,null,1]

Example details omitted.

Constraints

The number of nodes in the tree is in the range [1, 104].
The depth of the tree is in the range [1, 104].
-100 <= Node.val <= 100
-105 <= val <= 105
1 <= depth <= the depth of tree + 1

Solution Approach

DFS Approach

In a Depth-First Search (DFS) approach, traverse the tree while keeping track of the current depth. Once the target depth is reached, insert new nodes at that level by adjusting the left and right children of the current node.

BFS Approach

The Breadth-First Search (BFS) approach uses a queue to traverse the tree level by level. When reaching the target depth, add the new nodes at the appropriate level and adjust parent-child links.

Optimized Approach

Combine DFS and BFS strategies with a focus on reducing unnecessary operations, such as only adding new nodes when required. This ensures optimal space and time complexity.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time and space complexities vary depending on the chosen approach. A DFS approach typically has a time complexity of O(N) where N is the number of nodes, and space complexity can range from O(H) to O(N) based on the recursion stack or queue size. BFS usually has O(N) time and O(N) space complexity due to the level-order traversal and queue operations.

What Interviewers Usually Probe

Ability to identify the most efficient traversal method for tree modifications.
Proficiency in managing binary-tree relationships while maintaining integrity.
Strong understanding of binary-tree operations and performance considerations.

Common Pitfalls or Variants

Common pitfalls

Incorrectly handling depth 1 insertion by not adjusting the root.
Failing to manage left and right children properly when inserting nodes.
Not accounting for edge cases where the depth exceeds the tree's current depth.

Follow-up variants

Modify the tree at a given depth with multiple values for nodes instead of a single value.
Allow dynamic depth changes while performing insertions at various levels.
Handle unbalanced trees by adding rows at varying depths.

FAQ

How does binary-tree traversal help solve the Add One Row to Tree problem?

Binary-tree traversal helps you reach the target depth and allows you to add new nodes while maintaining the tree's structure, ensuring efficient solutions.

What are common mistakes when adding a row to a binary tree?

Common mistakes include improperly handling the root insertion or mismanaging the left and right children of parent nodes at the target depth.

How do Depth-First Search and Breadth-First Search differ in this problem?

DFS focuses on recursion and depth tracking, while BFS uses a queue to perform level-order traversal, each offering unique advantages based on the tree's structure.

What is the optimal approach for the Add One Row to Tree problem?

The optimal approach is typically a BFS or DFS that efficiently tracks depth and modifies the tree with minimal overhead, depending on space and time complexity requirements.

Can this problem be solved with a recursive approach?

Yes, a recursive DFS approach can solve this problem by keeping track of depth and inserting nodes when the target depth is reached.

terminal

Solution

Solution 1

#### Python3

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# Definition for a binary tree node.
# 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]:
        def dfs(root, d):
            if root is None:
                return
            if d == depth - 1:
                root.left = TreeNode(val, root.left, None)
                root.right = TreeNode(val, None, root.right)
                return
            dfs(root.left, d + 1)
            dfs(root.right, d + 1)

        if depth == 1:
            return TreeNode(val, root)
        dfs(root, 1)
        return root

Solution 2

#### Python3

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# Definition for a binary tree node.
# 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]:
        def dfs(root, d):
            if root is None:
                return
            if d == depth - 1:
                root.left = TreeNode(val, root.left, None)
                root.right = TreeNode(val, None, root.right)
                return
            dfs(root.left, d + 1)
            dfs(root.right, d + 1)

        if depth == 1:
            return TreeNode(val, root)
        dfs(root, 1)
        return root

Continue Practicing

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