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InorderPreAndSucc.cpp
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128 lines (108 loc) · 2.21 KB
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#include <iostream>
#include <vector>
using namespace std;
// Node class
class Node
{
public:
int data;
Node *left;
Node *right;
Node(int val)
{
data = val;
left = right = NULL;
}
};
// isert node
Node *insert(Node *root, int val)
{
// base case
if (root == NULL)
{
return new Node(val);
}
// left sub tree call
if (val < root->data)
{
root->left = insert(root->left, val);
}
// right sub tree call
else
{
root->right = insert(root->right, val);
}
return root;
}
// right Most In Left Subtree
Node *rightMostInLeftSubtree(Node *root)
{
Node *ans;
while (root != NULL)
{
ans = root;
root = root->right;
}
return ans;
}
// left Most In Right Subtree
Node *leftMostInRightSubtree(Node *root)
{
Node *ans;
while (root != NULL)
{
ans = root;
root = root->left;
}
return ans;
}
// find Predecessor & Successor in BST
vector<int> getPredSucc(Node *root, int key)
{
Node *curr = root; // current node
Node *pred = NULL; // Predecessor
Node *succ = NULL; // Successor
while (curr != NULL)
{
if (key < curr->data)
{
succ = curr;
curr = curr->left;
}
else if (key > curr->data)
{
pred = curr;
curr = curr->right;
}
else
{
if (curr->left != NULL)
{
// inorder pred
pred = rightMostInLeftSubtree(curr->left);
}
if (curr->right != NULL)
{
// inorder succ
succ = leftMostInRightSubtree(curr->right);
}
break;
}
}
return {pred->data, succ->data};
}
int main()
{
Node *root = new Node(6);
root->left = new Node(4);
root->right = new Node(8);
root->left->left = new Node(1);
root->left->right = new Node(5);
root->right->left = new Node(7);
root->right->right = new Node(9);
int key = 7;
vector<int> ans = getPredSucc(root, key);
cout << "Predecessor: " << ans[0] << endl;
cout << "Successor: " << ans[1] << endl;
return 0;
}