Populate Inorder Successor for all nodes in C++

In this problem, we are given a tree. The structure contains a pointer next. Our task is to populate this pointer with the inorder successor of the node.

struct node {
   int value;
   struct node* left;
   struct node* right;
   struct node* next;

All the next pointer are set to NULL and we have to set the pointer to the inorder successor of the node.

Inorder traversal − This traverses in the following form,

Left node -> root Node -> right node.

Inorder successor − the inorder successor is the node that comes after the current node is the inorder traversal of the tree.

Let’s take an example to understand the problem,

The in-order traversal is 7 8 3 5 9 1

Populating each node −

Next of 5 is 9
Next of 8 is 3
Next of 7 is 8
Next of 3 is 5
Next of 9 is 1

To solve this problem, we will traverse the tree but in a reverse in the order form. Then we will put the last visit element to the next of the number.


Program to show the implementation of our solution,

 Live Demo

using namespace std;
struct node {
   int data;
   node *left;
   node *right;
   node *next;
node* insertNode(int data){
   node* Node = new node();
   Node->data = data;
   Node->left = NULL;
   Node->right = NULL;
   Node->next = NULL;
void populateTree(node* pop){
   static node *next = NULL;
   if (pop){
      pop->next = next;
      next = pop;
void printNext(node * root) {
   node *ptr = root->left->left;
      cout<<"Next of "<<ptr->data<<" is ";
      cout<<(ptr->next? ptr->next->data: -1)<<endl;
      ptr = ptr->next;
int main() {
   node *root = insertNode(15);
   root->left = insertNode(99);
   root->right = insertNode(1);
   root->left->left = insertNode(76);
   root->left->right = insertNode(31);
   cout<<"Populating the Tree by adding inorder successor to the next\n";
   return 0;


Populating the Tree by adding inorder successor to the next

Next of 76 is 99
Next of 99 is 31
Next of 31 is 15
Next of 15 is 1
Next of 1 is -1

Updated on: 17-Apr-2020


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