Binary Tree Upside Down in C++

Suppose we have a binary tree where all the right nodes are either leaf nodes with a sibling otherwise empty, we have to flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. We have to return the new node.

So, if the input is like [1,2,3,4,5]

then the output will return the root of the binary tree [4,5,2,#,#,3,1]

To solve this, we will follow these steps −

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

• if node is not present, then −

  • return NULL

• child = left of node

• currSib = right of node

• node := left of sibling

• node := right of par

• if child and currSib is not present, then −

  • return node

• return solve(child, node, currSib)

• From the main method do the following −

• return solve(root, NULL, NULL)

Example 

Let us see the following implementation to get 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 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;
}
void inord(TreeNode *root){
   if(root != NULL){
      inord(root->left);
      cout << root->val << " ";
      inord(root->right);
   }
}
class Solution {
public:
   TreeNode* solve(TreeNode* node, TreeNode* par, TreeNode* sibling){
      if (!node || node->val == 0)
         return NULL;
      TreeNode* child = node->left;
      TreeNode* currSib = node->right;
      node->left = sibling;
      node->right = par;
      if (!child && !currSib)
         return node;
      return solve(child, node, currSib);
   }
   TreeNode* upsideDownBinaryTree(TreeNode* root) {
      return solve(root, NULL, NULL);
   }
};
main(){
   Solution ob;
   vector<int< v = {1,2,3,4,5};
   TreeNode *root = make_tree(v);
   inord(ob.upsideDownBinaryTree(root));
}

Input

[1,2,3,4,5]

Output

[4,5,2,null,null,3,1]