Maximum Element of Binary Tree added

Author: neha030

Data Structures/Trees/Maximum_Element_of_Binary_Tree.cpp

@@ -0,0 +1,153 @@
+/*
+Given a Binary Tree , Print the value of maximum node key 
+
+Argument/Return Type
+Input of total no.of nodes is taken
+Input of key values of nodes of tree are taken in level order form 
+Incase of a null node , -1 is taken as input
+A function which returns the maximum element 
+If the root is NULL it prints -1
+*/
+
+#include <bits/stdc++.h> 
+using namespace std;
+
+//Define Node as structure
+struct Node 
+{
+    int key;
+    Node* left;
+    Node* right;
+};
+ 
+// Function to create a node with 'value' as the data stored in it. 
+// Both the children of this new Node are initially null.
+struct Node* newNode(int value)
+{
+    Node* n = new Node;
+    n->key = value;
+    n->left = NULL;
+    n->right = NULL;
+    return n;
+}
+
+// Function to build tree with given input
+struct Node* createTree(vector<int>v)
+{
+    int n=v.size();
+    if(n==0) 
+      return NULL;
+    vector<struct Node* >a(n);
+    //Create a vector of individual nodes with given node values
+    for(int i=0;i<n;i++)  
+    {
+        //If the data is -1 , create a null node
+        if(v[i]==-1)  
+          a[i] = NULL;
+        else 
+          a[i] = newNode(v[i]);
+    }
+    //Interlink all created nodes to create a tree
+    //Use two pointers using int to store indexes
+    //One to keep track of parent node and one for children nodes
+    for(int i=0,j=1;j<n;i++) 
+    {
+        //If the parent node is NULL , advance children pointer twice
+        if(!a[i])
+        {
+          j=j+2;
+          continue;
+        } 
+        //Connect the two children nodes to parent node
+        //First left and then right nodes
+        a[i]->left = a[j++];
+        if(j<n) 
+          a[i]->right = a[j++];
+    }
+    return a[0];
+}
+
+//Function to return maximum value of root key of the given tree
+int FindMax(struct Node* root)
+{
+
+    // If root is NULL , return -1
+    if (root == NULL)
+        return -1;
+    //Initialise a maximum variable with -1 value 
+    // keep track of it traversing the tree in Pre Order Fashion
+    int maximum=-1;
+
+    // Create an empty stack and push root to it
+    stack<struct Node*> stack;
+ 
+    while (1) 
+    {
+        while(root)
+        {
+            //keep visiting left branch , untill we reach extreme left node
+            //During each visit , keep updating the maximum value
+            maximum=max(maximum,root->key);
+            stack.push(root);
+            root=root->left;
+        }
+        if(stack.empty())
+           break;
+        // Now consider the most recently visited node
+        root=stack.top();
+        stack.pop();
+        //and run Pre Order traversal for its right child 
+        root=root->right;
+        //repeat the process till there are no more nodes left
+    }
+    return  maximum;
+}
+ 
+// Driver code
+int main()
+{
+
+    int n;
+    cout<<"Enter total no.of nodes of the input Tree ( including NULL nodes ) : ";
+    cin>>n;
+
+    vector<int>v(n);
+    cout<<"Enter value of each node of the tree in level order ( if a node is NULL , enter -1 ) with spaces"<<endl;
+    for(int i=0;i<n;i++)
+    {
+        cin>>v[i]; //store the input values in a vector
+    }
+
+    //create the tree using input node values  
+    struct Node* root=createTree(v); 
+
+    //Call the function and print the result
+    cout<<"The maximum value of the node key of given tree is "<<FindMax(root);
+    
+    return 0; 
+}
+    
+/*
+if node is NULL , -1 is entered as it's key
+Sample Test Case  
+Input Binary Tree :                   
+               10                                
+           /         \                          
+         11              12                    
+     /       \        /   \               
+    5       NULL     6     13             
+   / \      /  \    / \   /  \            
+  4 NULL  NULL NULL 8  9 10  NULL         
+Input : 
+Enter total no.of nodes of the input Tree ( including NULL nodes ) : 15
+Enter value of each node of the tree in level order ( if a node is NULL , enter -1 ) with spaces
+10 11 12 5 -1 6 13 4 -1 -1 -1 8 9 10 -1
+
+Output :
+The maximum value of the node key of given tree is 13
+
+Time Complexity : O(n) 
+   Where n is the no.of nodes 
+Space Complexity : O(n) 
+   Where n is the no.of nodes
+*/