Solution Approach

The solution implements a recursive DFS approach with post-order traversal to find the longest consecutive sequence in the binary tree.

Core Algorithm Structure:

The implementation uses a helper function dfs(root) that returns the length of the longest consecutive sequence starting from the current node and going downward.

Step-by-step Implementation:

1. Base Case: If the current node is None, return 0 since there's no sequence to form.

2. Recursive Calls: First, recursively compute the longest consecutive sequences from both children:

   l = dfs(root.left) + 1 r = dfs(root.right) + 1

   We add 1 to include the current node in the potential sequence length.

3. Validate Consecutive Property: Check if each child actually continues the consecutive sequence:

   if root.left and root.left.val - root.val != 1:  l = 1 if root.right and root.right.val - root.val != 1:  r = 1

   If a child exists but its value isn't exactly one more than the current node's value, reset that path's length to 1 (only the current node).

4. Calculate Local Maximum: Find the longest consecutive sequence starting from the current node:

   t = max(l, r)

5. Update Global Maximum: Use the nonlocal keyword to update the global answer:

   nonlocal ans ans = max(ans, t)

   This ensures we track the longest sequence found anywhere in the tree.

6. Return Value: Return t to the parent call, which represents the longest consecutive sequence starting from this node.

Main Function:

• Initialize ans = 0 to track the global maximum
• Call dfs(root) to traverse the entire tree
• Return the final answer

The time complexity is O(n) where n is the number of nodes, as we visit each node exactly once. The space complexity is O(h) where h is the height of the tree, due to the recursion call stack.

Example Walkthrough

Let me walk through a small example to illustrate how this solution works.

Consider this binary tree:

 1  / \  3 2  / \  2 3

Step-by-step DFS traversal (post-order):

1. Visit leftmost leaf (value = 2):

   • dfs(node_2) at bottom left
   • Both children are None, so l = 0 + 1 = 1, r = 0 + 1 = 1
   • No children exist, so consecutive checks are skipped
   • t = max(1, 1) = 1
   • Update global: ans = max(0, 1) = 1
   • Return 1 to parent

2. Visit node with value = 3 (left child of root):

   • dfs(node_3)
   • Left child returned 1, so l = 1 + 1 = 2
   • Right child is None, so r = 0 + 1 = 1
   • Check left child: 2 - 3 = -1 ≠ 1, so reset l = 1
   • t = max(1, 1) = 1
   • Global remains: ans = max(1, 1) = 1
   • Return 1 to parent

3. Visit rightmost leaf (value = 3):

   • dfs(node_3) at bottom right
   • Both children are None, so l = 1, r = 1
   • t = 1
   • Global remains: ans = max(1, 1) = 1
   • Return 1 to parent

4. Visit node with value = 2 (right child of root):

   • dfs(node_2)
   • Left child is None, so l = 1
   • Right child returned 1, so r = 1 + 1 = 2
   • Check right child: 3 - 2 = 1 ✓, so r stays 2
   • t = max(1, 2) = 2
   • Update global: ans = max(1, 2) = 2
   • Return 2 to parent

5. Visit root (value = 1):

   • dfs(node_1)
   • Left child returned 1, so l = 1 + 1 = 2
   • Right child returned 2, so r = 2 + 1 = 3
   • Check left child: 3 - 1 = 2 ≠ 1, so reset l = 1
   • Check right child: 2 - 1 = 1 ✓, so r stays 3
   • t = max(1, 3) = 3
   • Update global: ans = max(2, 3) = 3
   • Return 3

Final answer: 3

The longest consecutive sequence is 1 → 2 → 3 (root to right child to right grandchild).

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 binary tree. Each node is visited exactly once during the traversal. At each node, the algorithm performs constant time operations:

• Checking if the node is null
• Recursive calls to left and right children
• Comparing values with children (if they exist)
• Updating local and global maximum values

Since we visit each of the n nodes exactly once and perform O(1) work at each node, the overall time complexity is O(n).

Space Complexity: O(h) where h is the height of the binary tree.

The space complexity is determined by the recursive call stack. In the worst case:

• For a skewed tree (essentially a linked list), the height h = n, resulting in O(n) space complexity
• For a balanced tree, the height h = log(n), resulting in O(log n) space complexity

The algorithm uses a constant amount of extra space for variables (l, r, t, and the global ans), but this doesn't affect the overall space complexity which is dominated by the recursion stack depth.

Therefore, the space complexity is O(h), which ranges from O(log n) in the best case (balanced tree) to O(n) in the worst case (skewed tree).

Common Pitfalls

Pitfall 1: Incorrectly Handling the Sequence Direction

The Problem: A common mistake is misunderstanding how the consecutive sequence should flow. Some developers might incorrectly check if the parent's value is one more than the child's value (root.val - child.val == 1) instead of checking if the child's value is one more than the parent's value.

Incorrect Implementation:

# WRONG: Checking parent > child by 1 if root.left and root.val - root.left.val != 1:  left_length = 1

Correct Implementation:

# CORRECT: Checking child > parent by 1 if root.left and root.left.val - root.val != 1:  left_length = 1

Pitfall 2: Forgetting to Add 1 After Recursive Calls

The Problem: When calculating the consecutive sequence length, developers might forget to add 1 to include the current node in the path length, leading to off-by-one errors.

Incorrect Implementation:

# WRONG: Forgetting to include current node left_length = dfs(node.left) right_length = dfs(node.right)

Correct Implementation:

# CORRECT: Adding 1 to include current node left_length = dfs(node.left) + 1 right_length = dfs(node.right) + 1

Pitfall 3: Not Checking for Child Existence Before Accessing Values

The Problem: Attempting to access the val attribute of a child node without first checking if the child exists will cause an AttributeError when the child is None.

Incorrect Implementation:

# WRONG: May cause AttributeError if node.left is None if node.left.val - node.val != 1:  left_length = 1

Correct Implementation:

# CORRECT: Check existence first if node.left and node.left.val - node.val != 1:  left_length = 1

Pitfall 4: Returning Wrong Value from DFS Function

The Problem: Some might mistakenly return the global maximum from the DFS function instead of the local maximum starting from the current node. This breaks the recursive logic as parent nodes need to know the longest path starting from their children, not the global maximum.

Incorrect Implementation:

def dfs(node):  # ... calculation logic ...  nonlocal max_length  max_length = max(max_length, current_max)  return max_length # WRONG: Returns global max

Correct Implementation:

def dfs(node):  # ... calculation logic ...  nonlocal max_length  max_length = max(max_length, current_max)  return current_max # CORRECT: Returns local max from current node

Pitfall 5: Initializing Answer Variable Incorrectly

The Problem: Initializing the answer to 1 instead of 0 can cause issues when the tree is empty or when the actual maximum length is 0.

Incorrect Implementation:

max_length = 1 # WRONG: Assumes at least one node exists

Correct Implementation:

max_length = 0 # CORRECT: Handles empty tree case

Prevention Tips:

1. Always trace through the algorithm with a simple example to verify the logic
2. Consider edge cases like empty trees, single nodes, and paths that don't start from the root
3. Draw out the parent-child relationships and verify the direction of value comparison
4. Test with trees where no consecutive sequences exist (should return 1 if tree has nodes)