Replace each node in binary tree with the sum of its inorder predecessor and successor

3

Given a binary tree containing n nodes. The problem is to replace each node in the binary tree with the sum of its inorder predecessor and inorder successor.

Examples:

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

Output :        11
              /    \
             9      13
            / \    /  \
           2   3   4   3
                  
For 1:
Inorder predecessor = 5
Inorder successor  = 6
Sum = 11

For 4:
Inorder predecessor = 0
(as inorder predecessor is not present)
Inorder successor  = 2
Sum = 2

For 7:
Inorder predecessor = 3
Inorder successor  = 0
(as inorder successor is not present)
Sum = 3

Approach: Create an array arr. Store 0 at index 0. Now, store the inorder traversal of tree in the array arr. Then, store 0 at last index. 0’s are stored as inorder predecessor of leftmost leaf and inorder successor of rightmost leaf is not present. Now, perform inorder traversal and while traversing node replace node’s value with arr[i-1] + arr[i+1] and then increment i. In the beginning initialize i = 1. For an element arr[i], the values arr[i-1] and arr[i+1] are its inorder predecessor and inorder successor respectively.

// C++ implementation to replace each node 
// in binary tree with the sum of its inorder 
// predecessor and successor
#include <bits/stdc++.h>

using namespace std;

// node of a binary tree
struct Node {
    int data;
    struct Node* left, *right;
};

// function to get a new node of a binary tree
struct Node* getNode(int data)
{
    // allocate node
    struct Node* new_node = 
       (struct Node*)malloc(sizeof(struct Node));

    // put in the data;
    new_node->data = data;
    new_node->left = new_node->right = NULL;

    return new_node;
}

// function to store the inorder traversal
// of the binary tree in 'arr'
void storeInorderTraversal(struct Node* root, 
                                vector<int>& arr)
{
    // if root is NULL
    if (!root)
        return;

    // first recur on left child
    storeInorderTraversal(root->left, arr);

    // then store the root's data in 'arr'
    arr.push_back(root->data);

    // now recur on right child
    storeInorderTraversal(root->right, arr);
}

// function to replace each node with the sum of its
// inorder predecessor and successor
void replaceNodeWithSum(struct Node* root,
                        vector<int> arr, int* i)
{
    // if root is NULL
    if (!root)
        return;

    // first recur on left child
    replaceNodeWithSum(root->left, arr, i);

    // replace node's data with the sum of its
    // inorder predecessor and successor
    root->data = arr[*i - 1] + arr[*i + 1];

    // move 'i' to point to the next 'arr' element
    ++*i;

    // now recur on right child
    replaceNodeWithSum(root->right, arr, i);
}

// Utility function to replace each node in binary
// tree with the sum of its inorder predecessor 
// and successor
void replaceNodeWithSumUtil(struct Node* root)
{
    // if tree is empty
    if (!root)
        return;

    vector<int> arr;

    // store the value of inorder predecessor
    // for the leftmost leaf
    arr.push_back(0);

    // store the inoder traversal of the tree in 'arr'
    storeInorderTraversal(root, arr);

    // store the value of inorder successor
    // for the rightmost leaf
    arr.push_back(0);  

    // replace each node with the required sum
    int i = 1;
    replaceNodeWithSum(root, arr, &i);
}

// function to print the preorder traversal
// of a binary tree
void preorderTraversal(struct Node* root)
{
    // if root is NULL
    if (!root)
        return;

    // first print the data of node
    cout << root->data << " ";

    // then recur on left subtree
    preorderTraversal(root->left);

    // now recur on right subtree
    preorderTraversal(root->right);
}

// Driver program to test above
int main()
{
    // binary tree formation
    struct Node* root = getNode(1); /*         1        */
    root->left = getNode(2);        /*       /   \      */
    root->right = getNode(3);       /*     2      3     */
    root->left->left = getNode(4);  /*    /  \  /   \   */
    root->left->right = getNode(5); /*   4   5  6   7   */
    root->right->left = getNode(6);
    root->right->right = getNode(7);

    cout << "Preorder Traversal before tree modification:n";
    preorderTraversal(root);

    replaceNodeWithSumUtil(root);

    cout << "\nPreorder Traversal after tree modification:n";
    preorderTraversal(root);

    return 0;
}

Output:

Preorder Traversal before tree modification:
1 2 4 5 3 6 7
Preorder Traversal after tree modification:
11 9 2 3 13 4 3

Time Complexity: O(n)
Auxiliary Space: O(n)

This article is contributed by Ayush Jauhari. 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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