Given two Binary Trees, the task is to create a Maximum Binary Tree from the two given binary trees and print the Inorder Traversal of that tree.
What is the maximum Binary Tree?
The maximum binary is constructed in the following manner:
In the case of both the Binary Trees having two corresponding nodes, the maximum of the two values is considered as the node value of the Maximum Binary Tree.
If any of the two nodes is NULL and if the other node is not null, insert that value on that node of the Maximum Binary Tree.
Example:
Input:
Tree 1 Tree 2
3 5
/ \ / \
2 6 1 8
/ \ \
20 2 8
Output: 20 2 2 5 8 8
Explanation:
5
/ \
2 8
/ \ \
20 2 8
To construct the required Binary Tree,
Root Node value: Max(3, 5) = 5
Root->left value: Max(2, 1) = 2
Root->right value: Max(6, 8) = 8
Root->left->left value: 20
Root->left->right value: 2
Root->right->right value: 8
Input:
Tree 1 Tree 2
9 5
/ \ / \
2 6 1 8
/ \ \ \
20 3 2 8
Output: 20 2 3 9 8 8
Explanation:
9
/ \
2 8
/ \ \
20 3 8
Approach:
Follow the steps given below to solve the problem:
- Traverse both the trees using preorder traversal.
- If both the nodes are NULL, return. Otherwise, check for the following conditions:
- If both the nodes are not NULL then store the maximum between them as the node value of the Maximum Binary Tree.
- If only one of the node is NULL store the value of the non-NULL node as the node value of the Maximum Binary Tree.
- Recursively traverse the left subtrees.
- Recursively traverse the right subtrees
- Finally, return the root of the Maximum Binary Tree.
Below is the implementation of the above approach:
C++
// C++ program to find the Maximum // Binary Tree from two Binary Trees #include<bits/stdc++.h>using namespace std; // A binary tree node has data, // pointer to left child // and a pointer to right child class Node{ public:int data; Node *left, *right; Node(int data, Node *left, Node *right) { this->data = data; this->left = left; this->right = right; }}; // Helper method that allocates // a new node with the given data // and NULL left and right pointers. Node* newNode(int data) { Node *tmp = new Node(data, NULL, NULL); return tmp;} // Given a binary tree, print // its nodes in inordervoid inorder(Node *node) { if (node == NULL) return; // First recur on left child inorder(node->left); // Then print the data of node cout << node->data << " "; // Now recur on right child inorder(node->right); } // Method to find the maximum // binary tree from two binary treesNode* MaximumBinaryTree(Node *t1, Node *t2) { if (t1 == NULL) return t2; if (t2 == NULL) return t1; t1->data = max(t1->data, t2->data); t1->left = MaximumBinaryTree(t1->left, t2->left); t1->right = MaximumBinaryTree(t1->right, t2->right); return t1; } // Driver Code int main(){ /* First Binary Tree 3 / \ 2 6 / 20 */ Node *root1 = newNode(3); root1->left = newNode(2); root1->right = newNode(6); root1->left->left = newNode(20); /* Second Binary Tree 5 / \ 1 8 \ \ 2 8 */ Node *root2 = newNode(5); root2->left = newNode(1); root2->right = newNode(8); root2->left->right = newNode(2); root2->right->right = newNode(8); Node *root3 = MaximumBinaryTree(root1, root2); inorder(root3); } // This code is contributed by pratham76 |
Java
// Java program to find the Maximum// Binary Tree from two Binary Trees/* A binary tree node has data, pointer to left childand a pointer to right child */class Node { int data; Node left, right; public Node(int data, Node left, Node right) { this.data = data; this.left = left; this.right = right; } /* Helper method that allocates a new node with the given data and NULL left and right pointers. */ static Node newNode(int data) { return new Node(data, null, null); } /* Given a binary tree, print its nodes in inorder*/ static void inorder(Node node) { if (node == null) return; /* first recur on left child */ inorder(node.left); /* then print the data of node */ System.out.printf("%d ", node.data); /* now recur on right child */ inorder(node.right); } /* Method to find the maximum binary tree from two binary trees*/ static Node MaximumBinaryTree(Node t1, Node t2) { if (t1 == null) return t2; if (t2 == null) return t1; t1.data = Math.max(t1.data, t2.data); t1.left = MaximumBinaryTree(t1.left, t2.left); t1.right = MaximumBinaryTree(t1.right, t2.right); return t1; } // Driver Code public static void main(String[] args) { /* First Binary Tree 3 / \ 2 6 / 20 */ Node root1 = newNode(3); root1.left = newNode(2); root1.right = newNode(6); root1.left.left = newNode(20); /* Second Binary Tree 5 / \ 1 8 \ \ 2 8 */ Node root2 = newNode(5); root2.left = newNode(1); root2.right = newNode(8); root2.left.right = newNode(2); root2.right.right = newNode(8); Node root3 = MaximumBinaryTree(root1, root2); inorder(root3); }} |
Python3
# Python3 program to find the Maximum# Binary Tree from two Binary Trees ''' A binary tree node has data, pointer to left childand a pointer to right child '''class Node: def __init__(self, data, left, right): self.data = data self.left = left self.right = right ''' Helper method that allocates a new node with the given data and None left and right pointers. '''def newNode(data): return Node(data, None, None); ''' Given a binary tree, print its nodes in inorder'''def inorder(node): if (node == None): return; ''' first recur on left child ''' inorder(node.left); ''' then print the data of node ''' print(node.data, end=' '); ''' now recur on right child ''' inorder(node.right);''' Method to find the maximum binary tree from two binary trees'''def MaximumBinaryTree(t1, t2): if (t1 == None): return t2; if (t2 == None): return t1; t1.data = max(t1.data, t2.data); t1.left = MaximumBinaryTree(t1.left, t2.left); t1.right = MaximumBinaryTree(t1.right, t2.right); return t1;# Driver Codeif __name__=='__main__': ''' First Binary Tree 3 / \ 2 6 / 20 ''' root1 = newNode(3); root1.left = newNode(2); root1.right = newNode(6); root1.left.left = newNode(20); ''' Second Binary Tree 5 / \ 1 8 \ \ 2 8 ''' root2 = newNode(5); root2.left = newNode(1); root2.right = newNode(8); root2.left.right = newNode(2); root2.right.right = newNode(8); root3 = MaximumBinaryTree(root1, root2); inorder(root3);# This code is contributed by rutvik_56 |
C#
// C# program to find the Maximum // Binary Tree from two Binary Trees using System;// A binary tree node has data, // pointer to left child // and a pointer to right child class Node{ public int data; public Node left, right; public Node(int data, Node left, Node right) { this.data = data; this.left = left; this.right = right; } // Helper method that allocates // a new node with the given data // and NULL left and right pointers. static Node newNode(int data) { return new Node(data, null, null); } // Given a binary tree, print // its nodes in inorderstatic void inorder(Node node) { if (node == null) return; // first recur on left child inorder(node.left); // then print the data of node Console.Write("{0} ", node.data); // now recur on right child inorder(node.right); } // Method to find the maximum // binary tree from // two binary treesstatic Node MaximumBinaryTree(Node t1, Node t2) { if (t1 == null) return t2; if (t2 == null) return t1; t1.data = Math.Max(t1.data, t2.data); t1.left = MaximumBinaryTree(t1.left, t2.left); t1.right = MaximumBinaryTree(t1.right, t2.right); return t1; } // Driver Code public static void Main(String[] args) { /* First Binary Tree 3 / \ 2 6 / 20 */ Node root1 = newNode(3); root1.left = newNode(2); root1.right = newNode(6); root1.left.left = newNode(20); /* Second Binary Tree 5 / \ 1 8 \ \ 2 8 */ Node root2 = newNode(5); root2.left = newNode(1); root2.right = newNode(8); root2.left.right = newNode(2); root2.right.right = newNode(8); Node root3 = MaximumBinaryTree(root1, root2); inorder(root3); } } // This code is contributed by Amal Kumar Choubey |
Javascript
<script>// Javascript program to find the Maximum // Binary Tree from two Binary Trees // A binary tree node has data, // pointer to left child // and a pointer to right child class Node{ constructor(data) { this.data = data; this.left = null; this.right = null; }}// Helper method that allocates // a new node with the given data // and NULL left and right pointers. function newNode(data) { return new Node(data, null, null); } // Given a binary tree, print // its nodes in inorderfunction inorder(node) { if (node == null) return; // first recur on left child inorder(node.left); // then print the data of node document.write(node.data + " "); // now recur on right child inorder(node.right); } // Method to find the maximum // binary tree from // two binary treesfunction MaximumBinaryTree(t1, t2) { if (t1 == null) return t2; if (t2 == null) return t1; t1.data = Math.max(t1.data, t2.data); t1.left = MaximumBinaryTree(t1.left, t2.left); t1.right = MaximumBinaryTree(t1.right, t2.right); return t1; } // Driver Code /* First Binary Tree 3 / \ 2 6 / 20 */var root1 = newNode(3); root1.left = newNode(2); root1.right = newNode(6); root1.left.left = newNode(20); /* Second Binary Tree 5 / \ 1 8 \ \ 2 8 */var root2 = newNode(5); root2.left = newNode(1); root2.right = newNode(8); root2.left.right = newNode(2); root2.right.right = newNode(8); var root3 = MaximumBinaryTree(root1, root2); inorder(root3); </script> |
Output:
20 2 2 5 8 8
Time Complexity: O(N)
Auxiliary Space: O(N)
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