Finding diameter of a binary tree

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    

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;
  }
};