## Finding diameter of a binary tree

0 / 212 Problem statement:-

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.

Algorithm:-

The diameter of binary tree can be defined as max(Length of left subtree, Length of right subtree, Longest path between two nodes that passes through the root).

Steps:-

• Find the height of left subtree.
• Find the height of right subtree.
• Find the left diameter.
• Find the right diameter.
• Return the Maximum(Diameter of left subtree, Diameter of right subtree, Longest path between two nodes which passes through the root.)

Option 1:- O(N2) time complexity

```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public: int findDiameter(TreeNode * root, int & h) {
if (root == NULL) {
h = 0;
return 0;
}
int h1 = 0, h2 = 0;
int d1 = findDiameter(root -> left, h1);
int d2 = findDiameter(root -> right, h2);
h = max(h1, h2) + 1;
return max(h1 + h2, max(d1, d2));
}

int diameterOfBinaryTree(TreeNode * root) {
if (root == NULL) return 0;
int h;
return findDiameter(root, h);
}
};```

Option 2:- O(N) time complexity

In below example, we are doing 2 things in one pass , viz. finding the heights of left and right subtrees , computing the max of the left and right heights combined and returning the max of left and right subtree heights + 1 (for the root node)

```class Solution {
public: int diameterOfBinaryTree(TreeNode * root) {
int diameter = 0;
depth(root, diameter);
return diameter;
}
private: int depth(TreeNode * root, int & diameter) {
if (!root) return 0;
int left = depth(root->left, diameter);
int right = depth(root->right, diameter);
diameter = max(diameter, left + right);
return max(left, right) + 1;
}
};```