Merge Two Binary Trees by doing Node Sum (Recursive and Iterative)

2.5

Given two binary trees. We need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the non-null node will be used as the node of new tree.

Example:

Input: 
     Tree 1            Tree 2                  
       2                 3                             
      / \               / \                            
     1   4             6   1                        
    /                   \   \                      
   5                     2   7                  

Output: Merged tree:
         5
        / \
       7   5
      / \   \ 
     5   2   7

Note: The merging process must start from the root nodes of both trees.

Recursive Algorithm:

  1. Traverse the tree in Preorder fashion
  2. Check if both the tree nodes are NULL
    1. If not, then update the value
  3. Recur for left subtrees
  4. Recur for right subtrees
  5. Return root of updated Tree

C++

// C++ program to Merge 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 */
struct Node
{
    int data;
    struct Node *left, *right;
};

/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
Node *newNode(int data)
{
    Node *new_node = new Node;
    new_node->data = data;
    new_node->left = new_node->right = NULL;
    return new_node;
}

/* Given a binary tree, print its nodes in inorder*/
void inorder(Node * node)
{
    if (!node)
        return;

    /* first recur on left child */
    inorder(node->left);

    /* then print the data of node */
    printf("%d ", node->data);

    /* now recur on right child */
    inorder(node->right);
}

/* Function to merge given two binary trees*/
Node *MergeTrees(Node * t1, Node * t2)
{
    if (!t1)
        return t2;
    if (!t2)
        return t1;
    t1->data += t2->data;
    t1->left = MergeTrees(t1->left, t2->left);
    t1->right = MergeTrees(t1->right, t2->right);
    return t1;
}

// Driver code
int main()
{
    /* Let us construct the first Binary Tree
            1
          /   \
         2     3
        / \     \
       4   5     6
    */

    Node *root1 = newNode(1);
    root1->left = newNode(2);
    root1->right = newNode(3);
    root1->left->left = newNode(4);
    root1->left->right = newNode(5);
    root1->right->right = newNode(6);

    /* Let us construct the second Binary Tree
           4
         /   \
        1     7
       /     /  \
      3     2    6   */
    Node *root2 = newNode(4);
    root2->left = newNode(1);
    root2->right = newNode(7);
    root2->left->left = newNode(3);
    root2->right->left = newNode(2);
    root2->right->right = newNode(6);

    Node *root3 = MergeTrees(root1, root2);
    printf("The Merged Binary Tree is:\n");
    inorder(root3);
    return 0;
}

Java

// Java program to Merge Two Binary Trees

/* A binary tree node has data, pointer to left child
   and 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 merge given two binary trees*/
     static Node MergeTrees(Node t1, Node t2)
     {
         if (t1 == null)
             return t2;
         if (t2 == null)
             return t1;
         t1.data += t2.data;
         t1.left = MergeTrees(t1.left, t2.left);
         t1.right = MergeTrees(t1.right, t2.right);
         return t1;
     }
     
     // Driver method
     public static void main(String[] args)
     {
         /* Let us construct the first Binary Tree
                 1
               /   \
              2     3
             / \     \
            4   5     6
         */
      
         Node root1 = newNode(1);
         root1.left = newNode(2);
         root1.right = newNode(3);
         root1.left.left = newNode(4);
         root1.left.right = newNode(5);
         root1.right.right = newNode(6);
      
         /* Let us construct the second Binary Tree
                4
              /   \
             1     7
            /     /  \
           3     2    6   */
         Node root2 = newNode(4);
         root2.left = newNode(1);
         root2.right = newNode(7);
         root2.left.left = newNode(3);
         root2.right.left = newNode(2);
         root2.right.right = newNode(6);
      
         Node root3 = MergeTrees(root1, root2);
         System.out.printf("The Merged Binary Tree is:\n");
         inorder(root3);
     }
}
// This code is contributed by Gaurav Miglani


Output:

The Merged Binary Tree is:
7 3 5 5 2 10 12 

Complexity Analysis:

  • Time complexity : O(n)
    A total of n nodes need to be traversed. Here, n represents the minimum number of nodes from the two given trees.
  • Auxiliary Space : O(n)
    The depth of the recursion tree can go upto n in case of a skewed tree. In average case, depth will be O(logn).

Iterative Algorithm:

  1. Create a stack
  2. Push the root nodes of both the trees onto the stack.
  3. While the stack is not empty, perform following steps :
    1. Pop a node pair from the top of the stack
    2. For every node pair removed, add the values corresponding to the two nodes and update the value of the corresponding node in the first tree
    3. If the left child of the first tree exists, push the left child(pair) of both the trees onto the stack.
    4. If the left child of the first tree doesn’t exist, append the left child of the second tree to the current node of the first tree
    5. Do same for right child pair as well.
    6. If both the current nodes are NULL, continue with popping the next nodes from the stack.
  4. Return root of updated Tree
// C++ program to Merge 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 */
struct Node
{
    int data;
    struct Node *left, *right;
};

// Structure to store node pair onto stack
struct snode
{
    Node *l, *r;
};

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
Node *newNode(int data)
{
    Node *new_node = new Node;
    new_node->data = data;
    new_node->left = new_node->right = NULL;
    return new_node;
}

/* Given a binary tree, print its nodes in inorder*/
void inorder(Node * node)
{
    if (! node)
        return;

    /* first recur on left child */
    inorder(node->left);

    /* then print the data of node */
    printf("%d ", node->data);

    /* now recur on right child */
    inorder(node->right);
}

/* Function to merge given two binary trees*/

Node* MergeTrees(Node* t1, Node* t2)
{
    if (! t1)
        return t2;
    if (! t2)
        return t1;
    stack<snode> s;
    snode temp;
    temp.l = t1;
    temp.r = t2;
    s.push(temp);
    snode n;
    while (! s.empty())
    {
        n = s.top();
        s.pop();
        if (n.l == NULL|| n.r == NULL)
            continue;
        n.l->data += n.r->data;
        if (n.l->left == NULL)
            n.l->left = n.r->left;
        else
        {
            snode t;
            t.l = n.l->left;
            t.r = n.r->left;
            s.push(t);
        }
        if (n.l->right == NULL)
            n.l->right = n.r->right;
        else
        {
            snode t;
            t.l = n.l->right;
            t.r = n.r->right;
            s.push(t);
        }
    }
    return t1;
}

// Driver code
int main()
{
    /* Let us construct the first Binary Tree
            1
          /   \
         2     3
        / \     \
       4   5     6
    */
 
    Node *root1 = newNode(1);
    root1->left = newNode(2);
    root1->right = newNode(3);
    root1->left->left = newNode(4);
    root1->left->right = newNode(5);
    root1->right->right = newNode(6);
 
    /* Let us construct the second Binary Tree
           4
         /   \
        1     7
       /     /  \
      3     2    6   */
    Node *root2 = newNode(4);
    root2->left = newNode(1);
    root2->right = newNode(7);
    root2->left->left = newNode(3);
    root2->right->left = newNode(2);
    root2->right->right = newNode(6);
 
    Node *root3 = MergeTrees(root1, root2);
    printf("The Merged Binary Tree is:\n");
    inorder(root3);
    return 0;
}

Output:


The Merged Binary Tree is:
7 3 5 5 2 10 12 

Complexity Analysis:

  • Time complexity : O(n)
    A total of n nodes need to be traversed. Here, n represents the minimum number of nodes from the two given trees.
  • Auxiliary Space : O(n)
    The depth of the stack can go upto n in case of a skewed tree.

This article is contributed by Aakash Pal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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