# Replace Each Node in Binary Tree With The Sum Of Its Inorder Predecessor And Successor Using C++

C++ProgrammingServer Side Programming

#### C in Depth: The Complete C Programming Guide for Beginners

45 Lectures 4.5 hours

#### Practical C++: Learn C++ Basics Step by Step

Most Popular

50 Lectures 4.5 hours

#### Master C and Embedded C Programming- Learn as you go

66 Lectures 5.5 hours

We are given a binary tree, and we need to replace all the elements with the sum of its inorder predecessor and successor. Inorder is a traversed path in a graph that reads in the order of left node – root node – right node. The method adds the elements in left and right nodes of the parent node and replaces the value with the obtained sum.

Suppose we have a tree with the following formation and characters −

We can find and store the inorder of the tree in an array. After that, we can again do an inorder traversal, but we will replace the element with the indexes in our array this time.

Let us look at some input scenarios −

Assume the input given as the nodes of binary tree reading, [3 8 15 7 10 3 6 4 9], and the resultant binary tree is obtained as −

Input: [3 8 15 7 10 3 6 4 9]
Result—
Before: 3 8 15 7 10 3 6 4 9
After: 8 18 15 25 10 16 7 15 4


Assume another input given as the nodes of binary tree reading, [4 6 7 8 5 3 2 7 1], and the resultant binary tree is obtained as −

Input: [4 6 7 8 5 3 2 7 1]
Result—
Before: 8 6 2 5 7 4 7 1 3
After: 6 10 11 9 9 14 5 10 1


## Example

Following is the C++ implementation of the above method −

#include <iostream>
#include <vector>
using namespace std;
class Node {
public:
int value;
Node *left, *right;
Node(int value) {
this->value = value;
left = right = NULL;
}
};
void solve(Node* root, vector<int>& arr, int& index) {
if(root == NULL) return;
solve(root->left, arr, index);
index++;
root->value = arr[index-1] + arr[index+1];
solve(root->right, arr, index);
}
void storeInOrder(Node* root, vector<int>& arr) {
if(root == NULL) return;
storeInOrder(root->left, arr);
arr.push_back(root->value);
storeInOrder(root->right, arr);
}
void printInorder(Node* root) {
if(root == NULL) return;
printInorder(root->left);
cout << root->value << " ";
printInorder(root->right);
}
int main() {
Node* root = new Node(2);
root->left = new Node(7);
root->right = new Node(5);
root->left->left = new Node(2);
root->left->right = new Node(6);
root->right->right = new Node(9);
root->left->right->left = new Node(5);
root->left->right->right = new Node(11);
root->right->right->left = new Node(4);
cout << "Before: ";
printInorder(root);
cout << "\n";
vector<int> arr;
arr.push_back(0);
storeInOrder(root, arr);
arr.push_back(0);
int index = 0;
solve(root, arr, index);
cout << "After: ";
printInorder(root);
return 0;
}


## Output

Before: 2 7 5 6 11 2 5 4 9
After: 7 7 13 16 8 16 6 14 4


## Explanation

Node Inorder Predecessor Inorder Successor New Value
2(root) 11 5 16
7(root->left) 2 5 7
5(root->right) 2 4 6
2(root->left->left) 0(not available) 7 7
6(root->left->right) 5 11 16
9(root->right->right) 4 0(not available) 4
5(root->left->right->left) 7 6 13
11(root->left->right->right) 6 2 8
4(root->right->right->left) 5 9 14

## Conclusion

Storing the inorder and then finding the inorder of each element turn by turn is the key. We use the index as a backtrack to find out where the element was when inserted into the array, and then we compute the new in order of the tree. We traverse the tree and each node once, so the time complexity is O(n), and we store each element in the form of inorder, so the space complexity is O(n).

Updated on 10-Aug-2022 11:54:27