549. Binary Tree Longest Consecutive Sequence II

Problem Description

This problem provides us with the root of a binary tree and asks us to find the length of the longest consecutive path in that tree. The consecutive path we are looking for can be either increasing or decreasing in terms of node values, and every two consecutive nodes in this path should have values that differ by exactly one. To clarify, paths such as [1,2,3,4] (increasing) and [4,3,2,1] (decreasing) are valid. However, a path like [1,2,4,3] is invalid as the values do not differ by one. Additionally, it's important to note that the path does not need to follow parent-child relationships and could include a 'bounce', going from child to parent to another child (child-parent-child pattern).

Intuition

The intuition behind this solution lies in using a recursive depth-first search (DFS) algorithm to traverse the tree and compute the increasing and decreasing consecutive paths. For every node, there can be four kinds of consecutive paths:

1. Paths that only include this node.
2. Paths that start from this node and extend to any node in the left subtree.
3. Paths that start from this node and extend to any node in the right subtree.
4. Paths that go through this node, meaning they start from the left subtree, include this node, and go to the right subtree.

To track these paths, for each node, we calculate two values: incr and decr, which represent the lengths of the longest increasing and decreasing paths ending at this node, respectively.

When we visit a node, we check its children. If either child's value is one more than the current node's value, we increment incr; if it's one less, we increment decr. We take these two values from both children and use them to update the incr and decr values of the current node.

The magic happens by considering not only the deepest path from this node to each child but also the possibility of continuing the path by "bouncing" from one child to the other through the current node, which effectively can increase the overall length of the consecutive path.

As we compute the incr and decr values for each node, we keep track of the global maximum length (ans) found so far. To get the maximum consecutive path that can pass through a node, we add the incr and decr values of this node but subtract 1 because the node itself is counted twice (once in each path). After the DFS traverses the whole tree, ans will store the length of the longest consecutive path.

Learn more about Tree, Depth-First Search and Binary Tree patterns.

Solution Approach

The solution uses a recursive depth-first search (DFS) to explore the binary tree and calculate the length of the longest consecutive path. Here's a step-by-step walkthrough of the implementation:

1. Recursive Function Definition: A function dfs is defined, which takes a node of the tree as an argument and returns a tuple [incr, decr]. incr holds the maximum length of an increasing consecutive path ending at that node, and decr is the same for decreasing paths.

2. Base Case: When dfs encounters a None (indicating the node being visited is non-existent or a leaf's child), it returns [0, 0] as there are no consecutive paths beyond this point.

3. State Variables: The solution introduces a nonlocal variable ans to track the maximum length found during traversal.

4. Child Nodes Analysis: Each call to dfs considers the current node's left and right child nodes. For each child, the function computes i1/d1 and i2/d2 which are tuples returned by the recursive dfs call on the left and right children, respectively.

5. Update Increments/Decrements:

   • It then checks the value of the left child (if it exists), updating incr or decr based on whether the left child's value is one less or one more than the current node's value, respectively.
   • Similarly, checks are performed on the right child, updating incr and decr by comparing the values of the right child and the current node.

6. Global Maximum Update: After calculating the incr and decr for both the left and right children, ans is updated by taking the sum of the current node's incr and decr minus one — as the current node is counted in both incr and decr and should only be counted once.

7. Return Values: Finally, the function returns a tuple [incr, decr] for the current node, which signifies the maximum lengths of consecutive paths ending at this node (both increasing and decreasing).

8. Result: After dfs is called on the root, ans contains the length of the longest consecutive path in the tree, which is then returned as the result of the longestConsecutive function.

The algorithm effectively scans the entire tree only once, ensuring an efficient solution with a time complexity of O(n), where n is the number of nodes in the tree. By considering each node and its potential paths (both increasing and decreasing), as well as potential "bounces" between children, we ensure no possible paths are missed.

Example Walkthrough

Let's walk through an example to illustrate the solution approach. Consider the following binary tree:

1      3
2     / \
3    2   4
4   /     \
5  1       5

Here's how the recursive DFS would process this tree:

1. Starting at the Root (Node 3): The initial call to DFS is made on the root (node 3). At this point, ans is initialized to 0.

2. Recursive Calls: The dfs function will make recursive calls to the left (node 2) and right (node 4) children:

   • Left Child (Node 2):
     • The left child has the value 2, which is less by 1 than its parent (node 3), so for node 3, decr = 2.
     • It recursively calls dfs on node 2's left child (node 1).
     • Node 1 is also less by 1 than node 2, so now for node 2, decr = 2.
     • Node 1 has no children, so the recursive call would return [0, 0].
     • The decr from node 2 is now combined with node 3's decr to update ans, if necessary, to ans = decr_2 + decr_3 - 1.
   • Right Child (Node 4):
     • The right child has the value 4, which is more by 1 than its parent (node 3), so for node 3, incr = 2.
     • It recursively calls dfs on node 4's right child (node 5).
     • Node 5 is more by 1 than node 4, so for node 4, incr = 2.
     • The incr from node 4 is now combined with node 3's incr to update ans, if necessary, to ans = incr_4 + incr_3 - 1.

3. Update Global Maximum (ans): Since both left and right children of node 3 form consecutive sequences, we calculate the maximum sum of decr and incr.

   • From left child's decr (2 from node 2) and right child's incr (2 from node 4), we get 2 + 2 - 1 = 3 for node 3, which means including node 3, there is a path of length 3 that goes from node 5 to node 3 to node 2.

4. Finalizing the Result: After the full traversal,

   • We have computed incr and decr for all nodes.
   • We've also computed ans at each node, which is the maximum value obtained by adding incr and decr and subtracting 1 (to account for the current node being included in both sequences).
   • Since the tree was recursively traversed, ans holds the maximum length of the longest consecutive path after taking all nodes into consideration.

Based on our example, the longest consecutive path is 3 (which is the path [2, 3, 4] or [4, 3, 2]), and this is output by the longestConsecutive function as the DFS is executed from the root of the tree to all its children and their subsequent descendants.

Solution Implementation

Time and Space Complexity

Time Complexity

The provided code executes a depth-first search (DFS) on a binary tree. For every node, it calculates the longest consecutive sequence that can be formed both increasing and decreasing. The decision to increment or decrement the consecutive count, or to reset it, is done at every node. This leads to each node being visited exactly once.

Therefore, the time complexity of the DFS is O(N), where N is the number of nodes in the tree. This is because the function processes each node a single time without revisiting anything.

Space Complexity

The space complexity of the algorithm includes the space used by the recursive call stack during the DFS traversal as well as the space used for storing the variables in each recursive call.

In the worst case, the height of the binary tree may be O(N) (in case of a skewed tree where each node has only one child), which would imply O(N) recursive calls stack space would be used. For balanced trees, the average case height would be O(log N) leading to an average case space complexity of O(log N).

Hence, the space complexity is O(N) in the worst case, but in the average case for a balanced tree, it would be O(log N).

The ans variable used to store the maximum length of the consecutive sequence doesn't significantly affect the space complexity as it is a single integer value.

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