Program to pruning a given binary tree in C++


Suppose we have a binary tree, where every node's value is either a 0 or a 1. We have to find the same tree where every subtree not containing a 1 has been deleted. So if the tree is like −

To solve this, we will follow these steps −

  • Define a recursive method solve(), this will take the node. the method will be like −

  • If node is null, then return null

  • left of node := solve(left of node)

  • right of node := solve(right of node)

  • if left of node is null and right of node is also null and node value is 0, then return null

  • return node

Let us see the following implementation to get better understanding −

Example

 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 inorder(TreeNode *root){
   if(root){
      inorder(root->left);
      cout << root->val << ", ";
      inorder(root->right);
   }
}
class Solution {
   public:
   TreeNode* pruneTree(TreeNode* node) {
      if(!node)return NULL;
      node->left = pruneTree(node->left);
      node->right = pruneTree(node->right);
      if(!node->left && !node->right && !node->val){
         return NULL;
      }
      return node;
   }
};
main(){
   TreeNode *root = new TreeNode(1);
   root->left = new TreeNode(1);
   root->right = new TreeNode(0);
   root->left->left = new TreeNode(1);
   root->left->right = new TreeNode(1);
   root->right->left = new TreeNode(0);
   root->right->right = new TreeNode(1);
   root->left->left->left = new TreeNode(0);
   Solution ob;
   inorder(ob.pruneTree(root));
}

Input

TreeNode *root = new TreeNode(1);
root−>left = new TreeNode(1);
root−>right = new TreeNode(0);
root−>left−>left = new TreeNode(1);
root−>left−>right = new TreeNode(1);
root−>right−>left = new TreeNode(0);
root−>right−>right = new TreeNode(1);
root−>left−>left−>left = new TreeNode(0);

Output

1, 1, 1, 1, 0, 1,

Updated on: 21-Oct-2020

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