Solution Approach

The implementation uses a recursive DFS function with a visited set to track nodes we've already seen during traversal. Let's walk through the key components:

Data Structure: We use a set called vis to store references to nodes we've visited. This allows O(1) lookup time to check if a node has been seen before.

The DFS Function: The recursive dfs(root) function handles each subtree:

1. Base Cases:

   • If root is None: We've reached a leaf's child, return None
   • If root.right in vis: We've found the defective node! Its right child points to an already-visited node, so we return None to remove this entire subtree

2. Mark as Visited: Add the current node to the vis set. This must happen before processing children to ensure we can detect invalid pointers.

3. Recursive Processing:

   root.right = dfs(root.right) root.left = dfs(root.left)

   Notice the order - we process the right child first, then the left. This ordering is crucial because:

   • The invalid pointer points to a node on the same level to the right
   • By processing right subtrees first, we're more likely to visit the target node through the correct path before encountering the invalid pointer
   • When we later reach the defective node, its right child will already be in vis

4. Return the Node: If the node passes all checks, we return it unchanged, preserving the tree structure.

Why This Works: The algorithm exploits the fact that in a proper tree, each node should be reachable through exactly one path from the root. The defective node breaks this rule by creating a second path to some node. By tracking visited nodes, we can detect when a child pointer leads to an already-discovered node - this can only happen at the defective node.

The beauty of this approach is that it combines detection and removal in a single traversal. When we detect the invalid node (by finding its right child in vis), we immediately return None, which effectively removes that node and its entire subtree from the tree structure. The parent of the defective node will receive None and update its child pointer accordingly, seamlessly removing the problematic branch.

Example Walkthrough

Let's walk through a small example to illustrate how the solution identifies and removes the defective node.

Consider this binary tree where node 3 is defective (its right child incorrectly points to node 5):

 1  / \  2 4  / / \  3 5 6  \_____|  (invalid pointer)

Initial State:

• vis = {} (empty set)
• Start DFS from root (node 1)

Step 1 - Process Node 1:

• Check: Is node 1 null? No
• Check: Is node 1's right child (4) in vis? No
• Add node 1 to vis: vis = {1}
• Recursively process right child (node 4) first

Step 2 - Process Node 4:

• Check: Is node 4 null? No
• Check: Is node 4's right child (6) in vis? No
• Add node 4 to vis: vis = {1, 4}
• Recursively process right child (node 6)

Step 3 - Process Node 6:

• Check: Is node 6 null? No
• Check: Is node 6's right child (null) in vis? No
• Add node 6 to vis: vis = {1, 4, 6}
• Process children (both null), return node 6

Step 4 - Back to Node 4, Process Left Child (5):

• Check: Is node 5 null? No
• Check: Is node 5's right child (null) in vis? No
• Add node 5 to vis: vis = {1, 4, 5, 6}
• Process children (both null), return node 5
• Node 4 processing complete, return node 4

Step 5 - Back to Node 1, Process Left Child (2):

• Check: Is node 2 null? No
• Check: Is node 2's right child (null) in vis? No
• Add node 2 to vis: vis = {1, 2, 4, 5, 6}
• Process right child (null), then left child (node 3)

Step 6 - Process Node 3 (The Defective Node!):

• Check: Is node 3 null? No
• Check: Is node 3's right child (5) in vis? YES! Node 5 is already in vis
• This is the defective node! Return null to remove it

Final Result: Node 2's left child becomes null, effectively removing node 3 from the tree:

 1  / \  2 4  / \  5 6

The key moment was in Step 6: when we checked node 3's right child, we found it was already visited. This could only happen if the pointer was invalid, confirming node 3 as defective. By returning null, we removed node 3 and its entire subtree from the final tree structure.

Time and Space Complexity

Time Complexity: O(n), where n is the number of nodes in the binary tree.

The algorithm performs a depth-first search (DFS) traversal of the tree. In the worst case, it visits every node exactly once before finding and removing the incorrect node. Each node is processed in constant time O(1) - checking if root.right is in the visited set, adding the current node to the visited set, and making recursive calls. Therefore, the total time complexity is O(n).

Space Complexity: O(n), where n is the number of nodes in the binary tree.

The space complexity consists of two components:

1. Visited Set (vis): In the worst case, the algorithm stores all nodes in the visited set before finding the incorrect node, requiring O(n) space.
2. Recursion Call Stack: The DFS traversal uses the call stack for recursion. In the worst case (a skewed tree), the maximum depth of recursion can be O(n), requiring O(n) space for the call stack.

Since both components can require O(n) space in the worst case, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Processing Children in Wrong Order

One of the most critical pitfalls is processing the left child before the right child. This would cause the algorithm to fail in many cases.

Why it's a problem: The defective node's incorrect pointer points to a node on the same level to its right. If we traverse left-to-right (processing left child first), we might visit the defective node before visiting its incorrectly-pointed-to target, meaning the target won't be in our visited set yet, and we'll miss detecting the defective node.

Incorrect approach:

# WRONG: Processing left before right node.left = dfs(node.left) # Processing left first node.right = dfs(node.right) # Processing right second

Correct approach:

# CORRECT: Process right before left node.right = dfs(node.right) # Process right first node.left = dfs(node.left) # Process left second

2. Checking Visited Status After Adding to Set

Another common mistake is adding the current node to the visited set before checking if its children are already visited.

Why it's a problem: If you add the node to visited set first, then check if node.right in visited_nodes, you might incorrectly flag valid parent-child relationships as problematic.

Incorrect approach:

def dfs(node):  if node is None:  return None   visited_nodes.add(node) # Adding to visited first   # WRONG: Now checking children after marking current as visited  if node.right in visited_nodes:  return None

Correct approach:

def dfs(node):  # Check the right child BEFORE adding current node to visited  if node is None or node.right in visited_nodes:  return None   visited_nodes.add(node) # Add to visited AFTER the check

3. Using Node Values Instead of Node References

A subtle but important pitfall is using node values to track visited nodes instead of node references.

Why it's a problem: Multiple nodes can have the same value in a binary tree. Using values would cause false positives when different nodes have identical values.

Incorrect approach:

visited_nodes = set()  def dfs(node):  if node is None or (node.right and node.right.val in visited_nodes):  return None   visited_nodes.add(node.val) # Storing values, not references

Correct approach:

visited_nodes = set()  def dfs(node):  if node is None or node.right in visited_nodes:  return None   visited_nodes.add(node) # Store the actual node reference

4. Forgetting to Update Parent's Child Pointers

While the provided solution handles this correctly, a common implementation mistake is detecting the defective node but forgetting to actually remove it from the tree structure.

Why it's a problem: Simply identifying the defective node isn't enough; we need to update its parent's pointer to exclude the defective subtree.

Incorrect approach:

def dfs(node):  if node is None:  return None   if node.right in visited_nodes:  # Just returning without updating parent's pointer  return   visited_nodes.add(node)  dfs(node.right) # Not capturing return value  dfs(node.left) # Not capturing return value  return node

Correct approach:

def dfs(node):  if node is None or node.right in visited_nodes:  return None # Return None to signal removal   visited_nodes.add(node)  node.right = dfs(node.right) # Update child pointers  node.left = dfs(node.left) # Update child pointers  return node

These pitfalls highlight the importance of understanding both the tree traversal order and the mechanism of tree modification through recursive pointer updates.