Preorder predecessor of a Node in Binary Tree

Given a binary tree and a node in the binary tree, find Preorder predecessor of the given node.
Examples:

Consider the following binary tree
              20            
           /      \         
          10       26       
         /  \     /   \     
       4     18  24    27   
            /  \
           14   19
          /  \
         13  15
Input :  4
Output : 10
Preorder traversal of given tree is 20, 10, 4, 
18, 14, 13, 15, 19, 26, 24, 27.

Input :  19
Output : 15

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.

  1. If the given node is root, then return NULL as preorder predecessor.
  2. If node is the left child of its parent or left child of parent is NULL, then return parent as its preorder predecessor.
  3. If node is the right child of its parent and left child of parent exists, then predecessor would be the rightmost node (max value) of the left subtree of parent.
  4. If node is the right child of its parent and the parent has no left child, then predecessor would be the parent node (max value).

C++

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// CPP program to find preorder predecessor of
// given node.
#include <iostream>
using namespace std;
  
struct Node {
    struct Node *left, *right, *parent;
    int value;
};
  
// Utility function to create a new node with
// given value.
struct Node* newNode(int value)
{
    Node* temp = new Node;
    temp->left = temp->right = temp->parent = NULL;
    temp->value = value;
    return temp;
}
  
Node* preorderPredecessor(Node* root, Node* n)
{
    // Root has no predecessor in preorder
    // traversal
    if (n == root)
        return NULL;
  
    // If given node is left child of its
    // parent or parent's left is empty, then 
    // parent is Preorder Predecessor.
    Node* parent = n->parent;
    if (parent->left == NULL || parent->left == n)
        return parent;
  
    // In all other cases, find the rightmost 
    // child in left substree of parent.
    Node* curr = parent->left;
    while (curr->right != NULL)
        curr = curr->right;
  
    return curr;
}
  
// Driver code
int main()
{
    struct 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;
  
    struct Node* res = preorderPredecessor(root, root->left->right->right);
    if (res) 
        printf("Preorder predecessor of %d is %d\n",
               root->left->right->right->value, res->value);    
    else 
        printf("Preorder predecessor of %d is NULL\n",
               root->left->right->right->value);    
  
    return 0;
}

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Java

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// Java program to find preorder predecessor of 
// given node. 
class GfG { 
  
static class Node { 
     Node left, right, parent; 
    int value; 
  
// Utility function to create a new node with 
// given value. 
static Node newNode(int value) 
    Node temp = new Node(); 
    temp.left = null;
    temp.right = null;
    temp.parent = null
    temp.value = value; 
    return temp; 
  
static Node preorderPredecessor(Node root, Node n) 
    // Root has no predecessor in preorder 
    // traversal 
    if (n == root) 
        return null
  
    // If given node is left child of its 
    // parent or parent's left is empty, then 
    // parent is Preorder Predecessor. 
    Node parent = n.parent; 
    if (parent.left == null || parent.left == n) 
        return parent; 
  
    // In all other cases, find the rightmost 
    // child in left substree of parent. 
    Node curr = parent.left; 
    while (curr.right != null
        curr = curr.right; 
  
    return curr; 
  
// 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 = preorderPredecessor(root, root.left.right.right); 
    if (res != null
        System.out.println("Preorder predecessor of " + root.left.right.right.value + " is " + res.value);     
    else
        System.out.println("Preorder predecessor of " + root.left.right.right.value + " is null"); 
  

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Python3

“””Python3 program to find preorder
predecessor of given node.”””

# A Binary Tree Node
# Utility function to create a new tree node
class newNode:

# Constructor to create a newNode
def __init__(self, data):
self.value = data
self.left = None
self.right = self.parent = None

def preorderPredecessor(root, n):

# Root has no predecessor in preorder
# traversal
if (n == root) :
return None

# If given node is left child of its
# parent or parent’s left is empty, then
# parent is Preorder Predecessor.
parent = n.parent
if (parent.left == None or
parent.left == n):
return parent

# In all other cases, find the rightmost
# child in left substree of parent.
curr = parent.left
while (curr.right != None) :
curr = curr.right

return curr

# 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 = preorderPredecessor(root, root.left.right.right)
if (res):
print(“Preorder predecessor of”,
root.left.right.right.value, “is”, res.value)
else:
print(“Preorder predecessor of”,
root.left.right.right.value, “is None”)

# This code is contributed
# by SHUBHAMSINGH10

C#

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// C# program to find preorder predecessor of 
// given node.
using System;
  
class GfG
  
    class Node
    
        public Node left, right, parent; 
        public int value; 
    
  
    // Utility function to create a new node with 
    // given value. 
    static Node newNode(int value) 
    
        Node temp = new Node(); 
        temp.left = null;
        temp.right = null;
        temp.parent = null
        temp.value = value; 
        return temp; 
    
  
    static Node preorderPredecessor(Node root, Node n) 
    
        // Root has no predecessor in preorder 
        // traversal 
        if (n == root) 
            return null
  
        // If given node is left child of its 
        // parent or parent's left is empty, then 
        // parent is Preorder Predecessor. 
        Node parent = n.parent; 
        if (parent.left == null || parent.left == n) 
            return parent; 
  
        // In all other cases, find the rightmost 
        // child in left substree of parent. 
        Node curr = parent.left; 
        while (curr.right != null
            curr = curr.right; 
  
        return curr; 
    
  
    // 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 = preorderPredecessor(root, root.left.right.right); 
        if (res != null
            Console.WriteLine("Preorder predecessor of " +
                               root.left.right.right.value +
                               " is " + res.value);     
        else
            Console.WriteLine("Preorder predecessor of "
                                root.left.right.right.value +
                                " is null"); 
    
}
  
// This code is contributed by PrinciRaj1992

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Output:

Preorder predecessor of 19 is 15

Time Complexity : O(h) where h is height of given Binary Tree
Auxiliary Space : O(1) since no use of arrays, stacks, queues.



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