Recover Binary Tree through it preorder and inorder traversal

/***
Given preorder and inorder traversal of a tree, construct the binary tree.
Example:
preorder = {7,10,4,3,1,2,8,11}
inorder = {4,10,3,1,7,11,8,2}
***/

#include<iostream>

using namespace std;

class node {
public:
       int data;
       node *left, *right;
       node(int val) {
              this->data = val;
              this->left = nullptr;
              this->right = nullptr;
       }
};

class binarytree {
public:
       binarytree(int *pre, int *in, int sz) {
              preorder = pre;
              inorder = in;
              size = sz;
       }
       node *getroot();
       void printBT();
private:
       node *constructBT(int &index, int begin, int end);
       int findindex(int val, int begin, int end);
       int *preorder;
       int *inorder;
       int size;
       node *root;  
};

int binarytree::findindex(int val, int begin, int end) {
       for(int i = begin; i<=end; i++) {
              if(inorder[i] == val) return i;
       }
       return -1;
}

node * binarytree::constructBT(int &index, int begin, int end) {    
       if(begin > end) {
              index--;
              return nullptr;
       }

       node *n = new node(preorder[index]);
       if(begin == end) return n;

       node *left, *right;
       int id = findindex(preorder[index], begin, end);
       index++;
       left = constructBT(index, begin, id-1);
       index++;
       right = constructBT(index, id+1, end);

       n->left = left;
       n->right = right;

       return n;    
}

node * binarytree::getroot(){
       int index = 0;
       root = constructBT(index, 0, size-1);
       return root;
}

void printBT(node *root) {
       if (root == nullptr) return;
      
       printBT(root->left);
       cout<<root->data<<" ";
       printBT(root->right);
}

void main() {
       int preorder[] = {7,10,4,3,1,2,8,11};
       int inorder[] = {4,10,3,1,7,11,8,2};
       int sz = sizeof(preorder) / sizeof(int);

       binarytree bt(preorder, inorder, sz);
       node *root = bt.getroot();
       node *p = root;
       printBT(p);

}

Comments

Post a Comment

Popular posts from this blog

Sudoku

/*** Sudoku Given a partially filled 9×9 2D array ‘grid[9][9]‘, the goal is to assign digits (from 1 to 9) to the empty cells so that every row,   column, and subgrid of size 3×3 contains exactly one instance of the digits from 1 to 9. Backtracking Algorithm Like all other Backtracking problems, we can solve Sudoku by one by one assigning numbers to empty cells.  Before assigning a number, we check whether it is safe to assign. We basically check that the same number  is not present in current row, current column and current 3X3 subgrid. After checking for safety,  we assign the number, and recursively check whether this assignment leads to a solution or not.  If the assignment doesn’t lead to a solution, then we try next number for current empty cell.  And if none of number (1 to 9) lead to solution, we return false. ***/ #include <iostream> using namespace std; class Sudoku { public :      ...
Read more

Longest prefix matching - trie

/*** Longest prefix matching Given a dictionary of words and an input string, find the longest prefix of the string which is also a word in dictionary. Example: Let the dictionary contains the following words: {are, area, base, cat, cater, children, basement} Below are some input/output examples: -------------------------------------- Input String            Output -------------------------------------- caterer                 cater basemexy                base child                   < Empty > ***/ #include <iostream> #include <string> using namespace std; struct node {        char val; ...
Read more

Climbing Stairs

You are climbing a stair case. It takes  n  steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note:  Given  n  will be a positive integer. class Solution { public:     int climbStairs(int n) {         if (n == 0) return 0;                  if (n == 1) return 1;                  // dp space complexity is O(n)         vector<int> dp(n);         dp[0] = 1;         dp[1] = 2;                  for(int i = 2; i<n; i++)         {             dp[i] = dp[i-1] + dp[i-2];         }                  return dp[ n-1 ]...
Read more