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tree.cpp
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tree.cpp
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#include <iostream>
#include <vector>
#include <queue>
#define THOUSAND 1000
using namespace std;
template <typename T>
class TreeNode {
public:
T data;
vector<TreeNode<T>*> children;
TreeNode(T data) {
this->data = data;
}
~TreeNode() {
for (int i = 0; i < children.size(); i++) {
delete children[i];
}
}
};
TreeNode<int> *makeTree(int arr[], int size) {
if(arr==nullptr || size<=0) return nullptr;
int i=0;
queue<TreeNode<int>*> q;
TreeNode<int> *root = new TreeNode<int>(arr[i++]);
q.push(root);
while(!q.empty())
{
TreeNode<int> *curr = q.front();
q.pop();
int child = arr[i++];
for(int j=0;j<child;j++)
{
TreeNode<int> *node = new TreeNode<int>(arr[i++]);
curr->children.push_back(node);
q.push(node);
}
}
return root;
}
void printLevelWise(TreeNode<int>* root) {
/* Given a generic tree, print the input tree in level wise order. ####For
* printing a node with data N, you need to follow the exact format:
* N:x1,x2,x3,...,xn
* wherer, N is data of any node present in the binary tree. x1, x2, x3,
* ...., xn are the children of node N. There is no space in between. You
* need to print all nodes in the level order form in different lines.
* Input: 10 3 20 30 40 2 40 50 0 0 0 0
* 10:20,30,40
* 20:40,50
* 30:
* 40:
* 40:
* 50:
* */
if(root==nullptr) return;
queue<TreeNode<int>*> q;
q.push(root);
while(!q.empty())
{
TreeNode<int> *curr = q.front();
q.pop();
cout << curr->data << ':';
int childCount = curr->children.size();
if(childCount)
{
cout << curr->children[0]->data;
q.push(curr->children[0]);
}
for(int i=1; i<childCount; i++)
{
cout << ',' << curr->children[i]->data;
q.push(curr->children[i]);
}
cout << endl;
}
}
int countOfNodes(TreeNode<int>* root) {
/* Given a generic tree, count and return the total number of nodes present
* in the given tree.*/
if(root==nullptr) return 0;
int result = 1; // root node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
result += countOfNodes(root->children[i]);
return result;
}
int sumOfNodes(TreeNode<int>* root) {
/* Given a generic tree, count and return the sum of all nodes present in the
* given tree.*/
if(root==nullptr) return 0;
int result = root->data; // root node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
result += sumOfNodes(root->children[i]);
return result;
}
TreeNode<int>* maxDataNode(TreeNode<int>* root) {
/* Given a generic tree, find and return the node with maximum data. You need
* to return the complete node which is having maximum data. Return null if
* tree is empty. */
if(root==nullptr) return nullptr;
TreeNode<int> *result = root; // root node
int max = root->data; // root node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
{
TreeNode<int> *temp = maxDataNode(root->children[i]);
if(temp->data > max)
{
max = temp->data;
result = temp;
}
}
return result;
}
void printNodesAtDepthK(TreeNode<int>* root, int k) {
/* Given a generic tree and an integer k, print all the nodes which are at
* depth k. Root is at depth 0. */
if(root==nullptr || k<0) return;
if(k==0) /* Base Case */
{
cout << root->data << ' ';
return;
}
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
printNodesAtDepthK(root->children[i], k-1);
}
int nodesGreaterThanX(TreeNode<int> *root, int x) {
/* Given a tree and an integer x, find and return number of Nodes which are
* greater than x.*/
if(root==nullptr) return 0;
int result = 0;
if(root->data>x) result++;
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
result += nodesGreaterThanX(root->children[i], x);
return result;
}
int numLeafNodes(TreeNode<int>* root) {
/* Given a generic tree, count and return the number of leaf nodes present in
* the given tree.*/
if(root==nullptr) return 0;
int childCount = root->children.size();
if(childCount==0) return 1; // Base case
int result = 0;
for(int i=0; i<childCount; i++)
result += numLeafNodes(root->children[i]);
return result;
}
void postOrder(TreeNode<int>* root) {
/* post Order Traversal recursive */
if(root==nullptr) return;
int result = root->data; // root node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
postOrder(root->children[i]);
cout << root->data << ' ';
}
bool containsX(TreeNode<int>* root, int x) {
/* Given a generic tree and an integer x, check if x is present in the given
* tree or not. Return true if x is present, return false otherwise. */
if(root==nullptr) return false;
if(root->data==x) return true;
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
if(containsX(root->children[i], x)) return true;
return false;
}
int sumNode(TreeNode<int> *root){
/* Called by maxSumNode */
if(root==nullptr) return 0;
int sum = root->data; // Calculate sumNode for root Node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
sum += root->children[i]->data;
return sum;
}
TreeNode<int>* maxSumNode(TreeNode<int> *root){
/* Given a tree, find and return the node for which sum of data of all its
* immideate children and the node itself is maximum. In the sum, data of node
* itself and data of immediate children is to be taken. */
/* Solution: We need to traverse all the nodes and calculate the Sum for each
* Node. Also, we can check if maximum Sum is greater than maximum calculated
* till now. Here we are using post Order Traversal but we may use any
* traversal technique. */
if(root==nullptr) return nullptr;
TreeNode<int> *result = root; // root node
int maxSum = sumNode(root); // Calculate sumNode for root Node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
{
TreeNode<int> *temp = maxSumNode(root->children[i]);
int sum = sumNode(temp);
if(sum > maxSum)
{
maxSum = sum;
result = temp;
}
}
return result;
}
bool isIdentical(TreeNode<int> *root1, TreeNode<int> * root2) {
/* Given two Generic trees, return true if they are structurally identical
* i.e. they are made of nodes with the same values arranged in the same
* way.*/
if(root1==nullptr)
{
if(root2==nullptr) return true;
else return false;
}
/* Both root1 and root2 are non null */
if(root1->data != root2->data) return false;
int childCount1 = root1->children.size();
int childCount2 = root2->children.size();
if(childCount1 != childCount2) return false;
for(int i=0; i<childCount1; i++)
if(!isIdentical(root1->children[i],root2->children[i]))
return false;
return true;
}
TreeNode<int>* nextLargerElement(TreeNode<int> *root, int n) {
/* Given a generic tree and an integer n. Find and return the node with next
* larger element in the Tree i.e. find a node with value just greater than
* n. */
/* Solution: We need to traverse all the nodes and check for next larger
* Node. Here we are using post Order Traversal but we may use any
* traversal technique. */
if(root==nullptr) return nullptr;
int childCount = root->children.size();
if(childCount==0) // Base Case
{
if(root->data>n) return root; // Found node with value greater than n
return nullptr;
}
TreeNode<int> *result = nullptr;
if(root->data>n) result=root; // Found node with value greater than n
for(int i=0; i<childCount; i++)
{
TreeNode<int> *temp = nextLargerElement(root->children[i], n);
if(temp==nullptr) continue;
if(result==nullptr || result->data>temp->data)
result = temp;
}
return result;
}
TreeNode<int>* nextLargerElement2(TreeNode<int> *root, int n) {
/* Given a generic tree and an integer n. Find and return the node with next
* larger element in the Tree i.e. find a node with value just greater than
* n. */
if(root==nullptr) return nullptr;
if(root->data>n) return root;
int childCount = root->children.size();
TreeNode<int> *result;
for(int i=0; i<childCount; i++)
if((result=nextLargerElement(root->children[i], n))!=nullptr)
return result;
return nullptr;
}
TreeNode <int>* secondLargest(TreeNode<int> *root) {
/* Given a generic tree, find and return the node with second largest value
* in given tree. Return NULL if no node with required value is present. */
if(root==nullptr) return nullptr;
int childCount = root->children.size();
if(childCount==0) return nullptr;
// we have atleast two nodes: root node and one child node
TreeNode<int> *largest = root, *secLargest=root->children[0];
if(largest->data<secLargest->data)
{
secLargest = root;
largest=root->children[0];
}
queue<TreeNode<int>*> q;
q.push(root);
while(!q.empty())
{
TreeNode<int> *curr = q.front();
q.pop();
childCount = curr->children.size();
for(int i=0; i<childCount; i++)
{
q.push(curr->children[i]);
if(curr->children[i]->data > secLargest->data)
{
if(curr->children[i]->data > largest->data)
{
secLargest = largest;
largest = curr->children[i];
}
else
{
secLargest = curr->children[i];
}
}
}
}
return secLargest;
}
void replaceWithDepthValueHelper(TreeNode<int> *root, int value){
/* In a given Generic Tree, replace each node with its depth value. You need
* to just update the data of each node, no need to return or print anything.
* */
if(root==nullptr) return;
root->data = value; // root node
int childCount = root->children.size();
for(int i=0; i<childCount; i++)
replaceWithDepthValueHelper(root->children[i],value+1);
}
void replaceWithDepthValue(TreeNode<int> *root){
/* In a given Generic Tree, replace each node with its depth value. You need
* to just update the data of each node, no need to return or print anything.
* */
replaceWithDepthValueHelper(root,0);
}
int getArray(int arr[], int size)
{
cin >> size;
// Read the array
for(int j=0;j<size;++j)
cin >> arr[j];
return size;
}
int main()
{
int arr[THOUSAND];
int n = getArray(arr, THOUSAND);
TreeNode<int> *root = makeTree(arr, n);
TreeNode<int> *temp = secondLargest(root);
cout << temp->data;
return 0;
}