Preorder Successor of a Node in Binary Tree
Given a binary tree and a node in the binary tree, find preorder successor of the given node. It may be assumed that every node has parent link.
Examples:
Consider the following binary tree
20
/ \
10 26
/ \ / \
4 18 24 27
/ \
14 19
/ \
13 15
Input : 4
Output : 18
Preorder traversal of given tree is 20, 10, 4,
18, 14, 13, 15, 19, 26, 24, 27.
Input : 19
Output : 26
A simple solution is to first store Preorder traversal of the given tree in an array then linearly search given node and print node next to it.
Time Complexity : O(n)
Auxiliary Space : O(n)
An efficient solution is based on below observations.
- If left child of given node exists, then the left child is preorder successor.
- If left child does not exist and given node is left child of its parent, then its sibling is its preorder successor.
- If none of above conditions are satisfied (left child does not exist and given node is not left child of its parent), then we move up using parent pointers until one of the following happens.
- We reach root. In this case, preorder successor does not exist.
- Current node (one of the ancestors of given node) is left child of its parent, in this case preorder successor is sibling of current node.
C++
// CPP program to find preorder successor of // a node in Binary Tree. #include <iostream> using namespace std; struct Node { struct Node *left, *right, *parent; int key; }; Node* newNode(int key) { Node* temp = new Node; temp->left = temp->right = temp->parent = NULL; temp->key = key; return temp; } Node* preorderSuccessor(Node* root, Node* n) { // If left child exists, then it is preorder // successor. if (n->left) return n->left; // If left child does not exist, then // travel up (using parent pointers) // until we reach a node which is left // child of its parent. Node *curr = n, *parent = curr->parent; while (parent != NULL && parent->right == curr) { curr = curr->parent; parent = parent->parent; } // If we reached root, then the given // node has no preorder successor if (parent == NULL) return NULL; return parent->right; } int main() { Node* root = newNode(20); root->parent = NULL; root->left = newNode(10); root->left->parent = root; root->left->left = newNode(4); root->left->left->parent = root->left; root->left->right = newNode(18); root->left->right->parent = root->left; root->right = newNode(26); root->right->parent = root; root->right->left = newNode(24); root->right->left->parent = root->right; root->right->right = newNode(27); root->right->right->parent = root->right; root->left->right->left = newNode(14); root->left->right->left->parent = root->left->right; root->left->right->left->left = newNode(13); root->left->right->left->left->parent = root->left->right->left; root->left->right->left->right = newNode(15); root->left->right->left->right->parent = root->left->right->left; root->left->right->right = newNode(19); root->left->right->right->parent = root->left->right; Node* res = preorderSuccessor(root, root->left->right->right); if (res) { printf("Preorder successor of %d is %d\n", root->left->right->right->key, res->key); } else { printf("Preorder successor of %d is NULL\n", root->left->right->right->key); } return 0; } |
chevron_right
filter_none
Java
// Java program to find preorder successor of // a node in Binary Tree. class Solution { static class Node { Node left, right, parent; int key; }; static Node newNode(int key) { Node temp = new Node(); temp.left = temp.right = temp.parent = null; temp.key = key; return temp; } static Node preorderSuccessor(Node root, Node n) { // If left child exists, then it is preorder // successor. if (n.left != null) return n.left; // If left child does not exist, then // travel up (using parent pointers) // until we reach a node which is left // child of its parent. Node curr = n, parent = curr.parent; while (parent != null && parent.right == curr) { curr = curr.parent; parent = parent.parent; } // If we reached root, then the given // node has no preorder successor if (parent == null) return null; return parent.right; } // Driver code public static void main(String args[]) { Node root = newNode(20); root.parent = null; root.left = newNode(10); root.left.parent = root; root.left.left = newNode(4); root.left.left.parent = root.left; root.left.right = newNode(18); root.left.right.parent = root.left; root.right = newNode(26); root.right.parent = root; root.right.left = newNode(24); root.right.left.parent = root.right; root.right.right = newNode(27); root.right.right.parent = root.right; root.left.right.left = newNode(14); root.left.right.left.parent = root.left.right; root.left.right.left.left = newNode(13); root.left.right.left.left.parent = root.left.right.left; root.left.right.left.right = newNode(15); root.left.right.left.right.parent = root.left.right.left; root.left.right.right = newNode(19); root.left.right.right.parent = root.left.right; Node res = preorderSuccessor(root, root.left.right.right); if (res != null) { System.out.printf("Preorder successor of %d is %d\n", root.left.right.right.key, res.key); } else { System.out.printf("Preorder successor of %d is null\n", root.left.right.right.key); } } } // This code is contributed by Arnab Kundu |
chevron_right
filter_none
Python3
""" Python3 program to find preorder successor of a node in Binary Tree.""" # A Binary Tree Node # Utility function to create a new tree node class newNode: # Constructor to create a new node def __init__(self, data): self.key = data self.left = None self.right = None self.parent=None def preorderSuccessor(root, n) : # If left child exists, then it is # preorder successor. if (n.left) : return n.left # If left child does not exist, then # travel up (using parent pointers) # until we reach a node which is left # child of its parent. curr = n parent = curr.parent while (parent != None and parent.right == curr): curr = curr.parent parent = parent.parent # If we reached root, then the given # node has no preorder successor if (parent == None) : return None return parent.right # Driver Code if __name__ == '__main__': root = newNode(20) root.parent = None root.left = newNode(10) root.left.parent = root root.left.left = newNode(4) root.left.left.parent = root.left root.left.right = newNode(18) root.left.right.parent = root.left root.right = newNode(26) root.right.parent = root root.right.left = newNode(24) root.right.left.parent = root.right root.right.right = newNode(27) root.right.right.parent = root.right root.left.right.left = newNode(14) root.left.right.left.parent = root.left.right root.left.right.left.left = newNode(13) root.left.right.left.left.parent = root.left.right.left root.left.right.left.right = newNode(15) root.left.right.left.right.parent = root.left.right.left root.left.right.right = newNode(19) root.left.right.right.parent = root.left.right res = preorderSuccessor(root, root.left.right.right) if (res) : print("Preorder successor of", root.left.right.right.key, "is", res.key) else: print("Preorder successor of", root.left.right.right.key,"is None") # This code is contributed # by SHUBHAMSINGH10 |
chevron_right
filter_none
C#
// C# program to find preorder successor of // a node in Binary Tree. using System; class GFG { public class Node { public Node left, right, parent; public int key; }; static Node newNode(int key) { Node temp = new Node(); temp.left = temp.right = temp.parent = null; temp.key = key; return temp; } static Node preorderSuccessor(Node root, Node n) { // If left child exists, then it is preorder // successor. if (n.left != null) return n.left; // If left child does not exist, then // travel up (using parent pointers) // until we reach a node which is left // child of its parent. Node curr = n, parent = curr.parent; while (parent != null && parent.right == curr) { curr = curr.parent; parent = parent.parent; } // If we reached root, then the given // node has no preorder successor if (parent == null) return null; return parent.right; } // Driver code public static void Main(String []args) { Node root = newNode(20); root.parent = null; root.left = newNode(10); root.left.parent = root; root.left.left = newNode(4); root.left.left.parent = root.left; root.left.right = newNode(18); root.left.right.parent = root.left; root.right = newNode(26); root.right.parent = root; root.right.left = newNode(24); root.right.left.parent = root.right; root.right.right = newNode(27); root.right.right.parent = root.right; root.left.right.left = newNode(14); root.left.right.left.parent = root.left.right; root.left.right.left.left = newNode(13); root.left.right.left.left.parent = root.left.right.left; root.left.right.left.right = newNode(15); root.left.right.left.right.parent = root.left.right.left; root.left.right.right = newNode(19); root.left.right.right.parent = root.left.right; Node res = preorderSuccessor(root, root.left.right.right); if (res != null) { Console.Write("Preorder successor of {0} is {1}\n", root.left.right.right.key, res.key); } else { Console.Write("Preorder successor of {0} is null\n", root.left.right.right.key); } } } // This code is contributed by 29AjayKumar |
chevron_right
filter_none
Output:
Preorder successor of 19 is 26
Time Complexity : O(h) where h is height of given Binary Tree
Auxiliary Space : O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
