993. Cousins in Binary Tree

 993. Cousins in Binary Tree

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

Two nodes of a binary tree are cousins if they have the same depth with different parents.

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

 

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

 

Constraints:

• The number of nodes in the tree is in the range [2, 100].
• 1 <= Node.val <= 100
• Each node has a unique value.
• x != y
• x and y are exist in the tree.

The solution to the above question

/**

 * 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:

   

    

    bool isCousins(TreeNode* root, int x, int y) {

        queue<TreeNode*>q;

        

        TreeNode* xp;

        int dpx=1;

        int dpy=1;

        TreeNode* yp;

        int depth=0;

        q.push(root);

        q.push(NULL);

        if(root->val==x){

                       xp=root;

                       dpx=depth;

                   }

                   else if(root->val==y){

                       yp=root;

                       dpy=depth;

                   }

        while(!q.empty()){

            TreeNode* f=q.front();

            if(f!=NULL){

               if(f->left){

                q.push(f->left);

                   if(f->left->val==x){

                       xp=f;

                       dpx=depth;

                   }

                   else if(f->left->val==y){

                       yp=f;

                       dpy=depth;

                   }

            } 

                if(f->right){

                q.push(f->right);

                   if(f->right->val==x){

                       xp=f;

                       dpx=depth;

                   }

                   else if(f->right->val==y){

                       yp=f;

                       dpy=depth;

                   }

            }

                q.pop();

                

            }

            else{

                

                depth++;

                q.pop();

                if(!q.empty()){

                    q.push(NULL);

                }

            }

            

        }

        if(xp!=yp && dpy==dpx){

            return true;

        }

        return false;

        

    }

};