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.

  1. If left child of given node exists, then the left child is preorder successor.
  2. If left child does not exist and given node is left child of its parent, then its sibling is its preorder successor.
  3. 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++

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// 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;
}

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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

Output:

Preorder successor of 19 is 26

Time Complexity : O(h) where h is height of given Binary Tree
Auxiliary Space : O(1)



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