<p>Given the <code>root</code> of a binary tree, return <em>the length of the longest path, where each node in the path has the same value</em>. This path may or may not pass through the root.</p>

<p><strong>The length of the path</strong> between two nodes is represented by the number of edges between them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/ex1.jpg" style="width: 450px; height: 238px;" />
<pre>
<strong>Input:</strong> root = [5,4,5,1,1,null,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The shown image shows that the longest path of the same value (i.e. 5).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/ex2.jpg" style="width: 450px; height: 238px;" />
<pre>
<strong>Input:</strong> root = [1,4,5,4,4,null,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The shown image shows that the longest path of the same value (i.e. 4).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
	<li>The depth of the tree will not exceed <code>1000</code>.</li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  int calculateLongest(TreeNode* root,int &longest_path){
     if(root == NULL){
        return 0;
     }
     int leftpath=calculateLongest(root->left,longest_path);
     int rightpath=calculateLongest(root->right,longest_path);

     int leftval=0,rightval=0;
     if(root->left != NULL && root->val==root->left->val){
        leftval+= leftpath + 1; //plus 1 for my(root) and left connection
     }
      if(root->right != NULL && root->val==root->right->val){
        rightval+= rightpath + 1; //plus 1 for my(root) and left connection
     }

     longest_path=max(longest_path,leftval+rightval);
     return max(leftval,rightval);
  }
    int longestUnivaluePath(TreeNode* root) {
        if(root == NULL){
            return 0;
        }
        int longest_path=0;
        calculateLongest(root,longest_path);
        return longest_path;
    }
};
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