Finding diameter of a binary tree

/**
 * 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;
  }
};

 
  
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