int countNodes(TreeNode* root) { if (root==nullptr) return 0; int lh=findheightleft( root); int rh=findheightright( root); if (lh==rh) return (1<<lh)-1; return 1+countNodes(root->left)+countNodes(root->right);… int findheightright(TreeNode* node){ int height=0; while(node){ height++; node->right; } return height; }

The function CountNumberOfNodes below is a recursive algorithm to find the totalnumber of nodes in the binary tree rooted at node x. Fill in the four blanks to make thisfunction work correctly.CountNumberOfNodes(Node x){If x≠NIL← CountNumberOfNodes(x.left)← CountNumberOfNodes(x.right)ReturnElseReturn}

#include <stdio.h>#include <stdlib.h>// Structure for a binary tree nodestruct Node { int data; struct Node* left; struct Node* right;};… int value; scanf("%d", &value); root = insert(root, value); } int maxLevel = -1; printRightView(root, 0, &maxLevel); printf("\n"); return 0;}

public: void inord(TreeNode *root, vector <int> inorder){ if(!root) return; if(root->left) inord(root->left, inorder); inorder.push_back(root->val); if(root->right) inord(root->right, inorder); } vector<int> inorderTraversal(TreeNode* root) { // if(root==NULL) return {}; vector <int> inorder; inord(root, inorder); return inorder; }};

Problem statementsGiven a binary tree, find its height.Input FormatA single line of input contains string s which contains the node of the tree in level order fashion.'N' in the string represents NULL node.A string 5 1 1 2 N N 2 reprsents the tree 5 / \ 1 1 / \ 2 2Similarly a string 1 2 N 3 N represents the tree 1 / 2 / 3Constraints1 <= Number of nodes <= 10^51 <= Data of a node <= 10^5Output FormatPrint the height of the tree.Sample Input2 N 1 3 NSample Output3

Tree: Height of a Binary Tree