All Possible Full Binary Trees in C++

Suppose a full binary tree is a binary tree where each node has exactly 0 or 2 children. So we have to find a list of all possible full binary trees with N nodes. Each node of each tree in the answer must have node.val = 0. The returned trees can be in any order. So if the input is 7, then the trees are −
To solve this, we will follow these steps −
Define a map m of integer type key and tree type value.
Define a map m of integer type key and tree type value.
define a method called allPossibleFBT(), this will take N as input
define a method called allPossibleFBT(), this will take N as input
is N is 1, then create a tree with one node whose value is 0, and return
is N is 1, then create a tree with one node whose value is 0, and return
if m has the key N, then return m[N]  Define an array called temp, and req := N – 1
if m has the key N, then return m[N]  Define an array called temp, and req := N – 1
for left in range 1 to req – 1right := req – leftif left = 2 or right = 2, then go for next iterationleftPart := allPossibleFBT(left), rightPart := allPossibleFBT(right)for j in range 0 to size of leftPart - 1for k in range 0 to size of rightPart – 1root := a new node with value 0left of root := leftPart[j], right of root := rightPart[k]insert root into ans
for left in range 1 to req – 1
right := req – left
right := req – left
if left = 2 or right = 2, then go for next iteration
if left = 2 or right = 2, then go for next iteration
leftPart := allPossibleFBT(left), rightPart := allPossibleFBT(right)
leftPart := allPossibleFBT(left), rightPart := allPossibleFBT(right)
for j in range 0 to size of leftPart - 1for k in range 0 to size of rightPart – 1root := a new node with value 0left of root := leftPart[j], right of root := rightPart[k]insert root into ans
for j in range 0 to size of leftPart - 1
for k in range 0 to size of rightPart – 1root := a new node with value 0left of root := leftPart[j], right of root := rightPart[k]insert root into ans
for k in range 0 to size of rightPart – 1
root := a new node with value 0
root := a new node with value 0
left of root := leftPart[j], right of root := rightPart[k]
left of root := leftPart[j], right of root := rightPart[k]
insert root into ans
insert root into ans
set m[N] := ans and return.
set m[N] := ans and return.
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 = right = NULL;
      }
};
void tree_level_trav(TreeNode*root){
   if (root == NULL) return;
      cout << "[";
   queue<TreeNode *> q;
   TreeNode *curr;
   q.push(root);
   q.push(NULL);
   while (q.size() > 1) {
      curr = q.front();
      q.pop();
      if (curr == NULL){
         q.push(NULL);
      } else {
            if(curr->left)
               q.push(curr->left);
            if(curr->right)
               q.push(curr->right);
            if(curr == NULL || curr->val == 0){
               cout << "null" << ", ";
            } else {
            cout << curr->val << ", ";
         }
      }
   }
   cout << "]"<<endl;
}
class Solution {
   public:
   map < int, vector <TreeNode*> > m;
   vector<TreeNode*> allPossibleFBT(int N) {
      if(N == 1){
         vector <TreeNode*> temp;
         TreeNode *n = new TreeNode(1);
         n->left = new TreeNode(0);
         n->right = new TreeNode(0);
         temp.push_back(n);
         return temp;
      }
      if(m.count(N))return m[N];
      vector <TreeNode*> ans;
      int required = N - 1;
      for(int left = 1; left < required; left++){
         int right = required - left;
         if(left == 2 || right == 2)continue;
         vector <TreeNode*> leftPart = allPossibleFBT(left);
         vector <TreeNode*> rightPart = allPossibleFBT(right);
         for(int j = 0; j < leftPart.size(); j++){
            for(int k = 0; k < rightPart.size(); k++){
               TreeNode* root = new TreeNode(1);
               root->left = leftPart[j];
               root->right = rightPart[k];
               ans.push_back(root);
            }
         }
      }
      return m[N] = ans;
   }
};
main(){
   vector<TreeNode*> v;
   Solution ob;
   v = (ob.allPossibleFBT(7)) ;
   for(TreeNode *t : v){
      tree_level_trav(t);
   }
}
7
[1, 1, 1, null, null, 1, 1, null, null, 1, 1, null, null, null, null, ]
[1, 1, 1, null, null, 1, 1, 1, 1, null, null, null, null, null, null, ]
[1, 1, 1, 1, 1, 1, 1, null, null, null, null, null, null, null, null, ]
[1, 1, 1, 1, 1, null, null, null, null, 1, 1, null, null, null, null, ]
[1, 1, 1, 1, 1, null, null, 1, 1, null, null, null, null, null, null, ]