//This is a C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Node structure of the tree
struct node1 {
int data;
struct node1* left;
struct node1* right;
};
// For creating a new node
struct node1* newNode(int data)
{
struct node1* node1 = (struct node1*)malloc(sizeof(struct
node1));
node1->data = data;
node1->left = NULL;
node1->right = NULL;
return node1;
};
// Shows structure for return type of
// function findPerfectBinaryTree
struct returnType {
// For storing if sub-tree is perfect or not
bool isPerfect;
// For storing if sub-tree is complete or not
bool isComplete;
// Indicates size of the tree
int size1;
// Root of biggest complete sub-tree
node1* rootTree;
};
// Shows helper function that returns height
// of the tree given size
int getHeight(int size1)
{
return ceil(log2(size1 + 1));
}
// Shows function to return the biggest
// complete binary sub-tree
returnType findCompleteBinaryTree(struct node1* root)
{
// Declaring returnType that
// needs to be returned
returnType rt1;
// If root is NULL then it is considered as both
// perfect and complete binary tree of size 0
if (root == NULL) {
rt1.isPerfect = true;
rt1.isComplete = true;
rt1.size1 = 0;
rt1.rootTree = NULL;
return rt1;
}
// Shows recursive call for left and right child
returnType lv1 = findCompleteBinaryTree(root->left);
returnType rv1 = findCompleteBinaryTree(root->right);
// CASE - A
// It has been seen that if left sub-tree is perfect and right is
// complete and there height is also same then sub-tree root
// is also complete binary sub-tree with size equal to
// sum of left and right subtrees plus one for current root
if (lv1.isPerfect == true && rv1.isComplete == true
&& getHeight(lv1.size1) == getHeight(rv1.size1)) {
rt1.isComplete = true;
// It has been seen that if right sub-tree is perfect then
// root is also perfect
rt1.isPerfect = (rv1.isPerfect ? true : false);
rt1.size1 = lv1.size1 + rv1.size1 + 1;
rt1.rootTree = root;
return rt1;
}
// CASE - B
// It has been seen if left sub-tree is complete and right is
// also perfect and the left height is greater than height of right by one
// then sub-tree root will be a complete binary sub-tree with the size equal to
// sum of left and right subtrees plus one for current root.
// But sub-tree cannot be perfect binary sub-tree.
if (lv1.isComplete == true && rv1.isPerfect == true
&& getHeight(lv1.size1) == getHeight(rv1.size1) + 1) {
rt1.isComplete = true;
rt1.isPerfect = false;
rt1.size1 = lv1.size1 + rv1.size1 + 1;
rt1.rootTree = root;
return rt1;
}
// CASE - C
// Otherwise this sub-tree cannot be a complete binary tree
// and simply return the biggest sized complete sub-tree
// found till now in the left or right sub-trees
rt1.isPerfect = false;
rt1.isComplete = false;
rt1.size1 = max(lv1.size1, rv1.size1);
rt1.rootTree = (lv1.size1 > rv1.size1 ? lv1.rootTree :
rv1.rootTree);
return rt1;
}
// Shows function to print the inorder traversal of the tree
void inorderPrint(node1* root)
{
if (root != NULL) {
inorderPrint(root->left);
cout << root->data << " ";
inorderPrint(root->right);
}
}
// Driver code
int main()
{
// Creating the tree
struct node1* root = newNode(50);
root->left = newNode(30);
root->right = newNode(60);
root->left->left = newNode(5);
root->left->right = newNode(20);
root->right->left = newNode(45);
root->right->right = newNode(70);
root->right->left->left = newNode(10);
// Getting the biggest sized complete binary sub-tree
struct returnType ans1 = findCompleteBinaryTree(root);
cout << "Size : " << ans1.size1 << endl;
// Printing the inorder traversal of the found sub-tree
cout << "Inorder Traversal : ";
inorderPrint(ans1.rootTree);
return 0;
}
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