Count Univalue Subtrees in C++

Suppose we have a binary tree; we have to count the number of uni-value subtrees. Here the Uni-value subtree indicates all nodes of the subtree have the same value.

So, if the input is like root = [5,1,5,5,5,null,5],

then the output will be 4

To solve this, we will follow these steps −

• Define a function solve(), this will take node,

• if node is empty, then −

  • return true

• left := solve(left of node)

• right := solve(right of node)

• if left is false or right is false, then −

  • return false

• if left of node is present and val of node is not equal to value of left of node, then −

  • return false

• if right of node is present and val of node is not equal to value of right of node, then −

  • return false

• (increase ret by 1)

• return true

• From the main method do the following −

• ret := 0

• solve(root)

• return ret

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class TreeNode{
public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data){
      val = data;
      left = NULL;
      right = NULL;
   }
};
void insert(TreeNode **root, int val){
   queue<TreeNode*> q;
   q.push(*root);
   while(q.size()){
      TreeNode *temp = q.front();
      q.pop();
      if(!temp->left){
         if(val != NULL)
            temp->left = new TreeNode(val);
         else
            temp->left = new TreeNode(0);
         return;
      }
      else{
         q.push(temp->left);
      }
      if(!temp->right){
         if(val != NULL)
            temp->right = new TreeNode(val);
         else
            temp->right = new TreeNode(0);
         return;
      }
      else{
         q.push(temp->right);
      }
   }
}
TreeNode *make_tree(vector<int< v){
   TreeNode *root = new TreeNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      insert(&root, v[i]);
   }
   return root;
}
class Solution {
public:
   int ret;
   bool solve(TreeNode* node){
      if (!node || node->val == 0)
         return true;
      bool left = solve(node->left);
      bool right = solve(node->right);
      if (!left || !right)
         return false;
      if (node->left && node->left->val != 0 && node->val != node->left->val)
         return false;
      if (node->right && node->right->val != 0 && node->val != node->right->val)
         return false;
      ret++;
      return true;
   }
   int countUnivalSubtrees(TreeNode* root){
      ret = 0;
      solve(root);
      return ret;
   }
};
main(){
   Solution ob;
   vector<int< v = {5,1,5,5,5,NULL,5};
   TreeNode *root = make_tree(v);
   cout << (ob.countUnivalSubtrees(root));
}

Input

{5,1,5,5,5,NULL,5}

Output

4