# diameter-of-binary-tree

## Problem

Diameter of Binary Tree

## Problem Description

``````Given a binary tree, you need to compute the length of the diameter of the tree.
The diameter of a binary tree is the length of the longest path between any two nodes in a tree.
This path may or may not pass through the root.

Example:
Given a binary tree

1
/ \
2   3
/ \
4   5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.``````

## Solution

From problem description, longest path mah or may not pass through the root.

For example:

The idea is to calculate the longest path from left subtree, and right subtree, and record longest path (left, right, (left + right)).

Using helper class Depth to record each node depth and max diameter. This way we can optimize the time complexity to `O(n)`.

1. Recursively find the height of left subtree

2. Recursively find the height of right subtree

3. Calculate max diameter max(left diameter, right diameter, (left.height + right.height))

4. Return max depth and max diameter.

### Complexity Analysis

Time Complexity: `O(n)`

• n - number of tree nodes.

### Code

``````class Solution {
public int diameterOfBinaryTree(TreeNode root) {
return getDepth(root).max;
}

private Depth getDepth(TreeNode node) {
if (node == null) return new Depth(0, 0);
Depth left = getDepth(node.left);
Depth right = getDepth(node.right);
// calculate diameter of current node
int max = Math.max(Math.max(left.max, right.max), left.depth + right.depth);
// calculate max depth
int depth = 1 + Math.max(left.depth, right.depth);
return new Depth(max, depth);
}

class Depth {
int max;
int depth;
public Depth(int max, int depth) {
this.max = max;
this.depth = depth;
}
}``````

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