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