Understanding the Problem
The diameter of a binary tree is defined as the length of the longest path between any two nodes in the tree. This path may or may not pass through the root. The "length" here is measured in terms of the number of edges between the nodes on the path.
For example, in the binary tree [1, 2, 3, 4, 5]
, one possible longest path is from node 4 to node 5. The diameter in this case is 3 (edges between 4-2, 2-1, and 1-3).
Step-by-Step Solution with Example
step 1: Understand what needs to be calculated
We need to find the **longest path** between any two nodes in the binary tree. This path can lie completely in the left or right subtree or pass through the root.
step 2: Define a helper function to calculate height
We recursively calculate the height of each node’s left and right subtree. The height of a node is the maximum number of edges on the longest path from the node to a leaf.
step 3: At each node, compute the potential diameter
At every node, we calculate the sum of the left and right subtree heights. This value represents the diameter passing through that node.
step 4: Keep track of the global maximum diameter
As we compute diameters through all nodes, we update a global variable if the current node's diameter is greater than the previous maximum.
step 5: Return the final maximum diameter
After traversing the entire tree, the global variable holds the length of the longest path, i.e., the diameter.
step 6: Example with tree [1, 2, 3, 4, 5]
Let's consider the binary tree:
1
/ 2 3
/ 4 5
- Height of 4 = 0
- Height of 5 = 0
- Height of 2 = max(0, 0) + 1 = 1
- Diameter at node 2 = height(4) + height(5) = 0 + 0 = 0 (wrong!)
- Actually, we include the edge count between nodes: so diameter at node 2 = 1 + 1 = 2
- Height of 3 = 0
- Diameter at root node 1 = height(2) + height(3) = 2 + 1 = 3
So, the maximum diameter is 3 (edges in path 4-2-1-3).
Edge Cases
Single Node Tree
If the tree has only one node like [1]
, there are no edges. So, the diameter is 0.
Empty Tree
If the tree is empty (root is null
), then the diameter is 0. This must be handled to avoid runtime errors.
Skewed Tree (like Linked List)
Example: [1, 2, null, 3, null, 4]
(a left-skewed tree). The diameter is 3 (edges between 1-2, 2-3, 3-4).
Uneven Subtrees
Example: [1, 2, 3, 4, null, null, 5]
. Even though one subtree is deeper, the longest path can still pass through intermediate nodes, not necessarily the root.
Balanced vs Unbalanced Trees
In a balanced tree, the diameter often passes through the root. In unbalanced trees, it might lie entirely within the deeper subtree.
Finally
The key idea is to recursively calculate subtree heights and use them to determine the longest path at each node. Even though we’re calculating heights, the goal is to keep track of the **maximum** left+right height combination seen so far.
This approach ensures that all possible paths are considered, including those not passing through the root. Handling edge cases like empty trees or skewed trees makes the solution robust and beginner-friendly.
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