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Preorder predecessor of a Node in Binary Tree

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  • Difficulty Level : Easy
  • Last Updated : 18 Aug, 2022
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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. 

Complexity Analysis:

  • 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).

Implementation:

C++




// 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 subtree 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;
}

Java




// 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 subtree 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");
 
}
}

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 subtree 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#




// 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 subtree 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

Javascript




<script>
 
// Javascript program to find preorder predecessor of
// given node.
class Node
{
    constructor()
    {
        this.left = null;
        this.right = null;
        this.parent = null;
        this.value = 0;
    }
}
 
// Utility function to create a new node with
// given value.
function newNode(value)
{
    var temp = new Node();
    temp.left = null;
    temp.right = null;
    temp.parent = null;
    temp.value = value;
    return temp;
}
 
function preorderPredecessor(root, 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.
    var parent = n.parent;
    if (parent.left == null || parent.left == n)
        return parent;
         
    // In all other cases, find the rightmost
    // child in left subtree of parent.
    var curr = parent.left;
    while (curr.right != null)
        curr = curr.right;
    return curr;
}
 
// Driver code
var 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;
var res = preorderPredecessor(root, root.left.right.right);
if (res != null)
    document.write("Preorder predecessor of " +
                       root.left.right.right.value +
                       " is " + res.value);    
else
    document.write("Preorder predecessor of " +
                        root.left.right.right.value +
                        " is null");
 
// This code is contributed by rrrtnx.
</script>

Output

Preorder predecessor of 19 is 15

Complexity Analysis:

  • 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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