<p>Given the <code>root</code> of a binary tree with unique values and the values of two different nodes of the tree <code>x</code> and <code>y</code>, return <code>true</code> <em>if the nodes corresponding to the values </em><code>x</code><em> and </em><code>y</code><em> in the tree are <strong>cousins</strong>, or </em><code>false</code><em> otherwise.</em></p>

<p>Two nodes of a binary tree are <strong>cousins</strong> if they have the same depth with different parents.</p>

<p>Note that in a binary tree, the root node is at the depth <code>0</code>, and children of each depth <code>k</code> node are at the depth <code>k + 1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/12/q1248-01.png" style="width: 304px; height: 270px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4], x = 4, y = 3
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/12/q1248-02.png" style="width: 334px; height: 266px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4,null,5], x = 5, y = 4
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/13/q1248-03.png" style="width: 267px; height: 258px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4], x = 2, y = 3
<strong>Output:</strong> false
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 100]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
	<li>Each node has a <strong>unique</strong> value.</li>
	<li><code>x != y</code></li>
	<li><code>x</code> and <code>y</code> are exist in the tree.</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:

   void findlevel(TreeNode* root, int x,int y,int &level_x ,int &level_y)
   {
      queue<TreeNode*> q;
      q.push(root);
      int curr_level=0;

      while(!q.empty())
      {
        int n=q.size();
        while(n--)
        {
            TreeNode*temp=q.front();
            q.pop();
            if(temp->val==x)
            {
                level_x=curr_level;
            }
            if(temp->val==y)
            {
                level_y=curr_level;
            }
            if(temp->left)
            {
                q.push(temp->left);
            }
            if(temp->right)
            {
                q.push(temp->right);
            }
        }

        if(level_x!=level_y)
        {
            break;// no need to further go as thier level are not same 
        }
        if(level_x!=-1 || level_y!=-1)// as 
        {
            break;
        }
        curr_level++;
      }

   }

   bool parent(TreeNode* root, int x, int y)
   {
    if(!root)
    {
        return false;
    }
    if(root->left&&root->right)
    {
        if(root->left->val==x&&root->right->val==y)
        {
            return true;
        }
        if(root->left->val==y&&root->right->val==x)
        {
            return true;
        }
    }

    return (parent(root->left,x,y)||parent(root->right,x,y));

   }




    bool isCousins(TreeNode* root, int x, int y) 
    {
        int level_x=-1;
        int level_y=-1;
        findlevel(root,x,y,level_x,level_y);
        // findlevel(root,y,level_y);

        if(level_x!=level_y)
        {
            return false;
        }
        
       return !parent(root,x,y);

    }
};
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