Lowest Common Ancestor in a Binary Tree | Set 2 (Using Parent Pointer)

Given values of two nodes in a Binary Tree, find the Lowest Common Ancestor (LCA). It may be assumed that both nodes exist in the tree.

For example, consider the Binary Tree in diagram, LCA of 10 and 14 is 12 and LCA of 8 and 14 is 8.

Let T be a rooted tree. The lowest common ancestor between two nodes n1 and n2 is defined as the lowest node in T that has both n1 and n2 as descendants (where we allow a node to be a descendant of itself). Source : Wikipedia.

We have discussed different approaches to find LCA in set 1. Finding LCA becomes easy when parent pointer is given as we can easily find all ancestors of a node using parent pointer.

Below are steps to find LCA.

1. Create an empty hash table.
2. Insert n1 and all of its ancestors in hash table.
3. Check if n2 or any of its ancestors exist in hash table, if yes return the first existing ancestor.

Below is C++ implementation of above steps.

C

// C++ program to find lowest common ancestor using parent pointer #include <bits/stdc++.h> using namespace std;    // A Tree Node struct Node {     Node *left, *right, *parent;     int key; };    // A utility function to create a new BST node Node *newNode(int item) {     Node *temp = new Node;     temp->key = item;     temp->parent = temp->left = temp->right = NULL;     return temp; }    /* A utility function to insert a new node with    given key in Binary Search Tree */ Node *insert(Node *node, int key) {     /* If the tree is empty, return a new node */     if (node == NULL) return newNode(key);        /* Otherwise, recur down the tree */     if (key < node->key)     {         node->left  = insert(node->left, key);         node->left->parent = node;     }     else if (key > node->key)     {         node->right = insert(node->right, key);         node->right->parent = node;     }        /* return the (unchanged) node pointer */     return node; }    // To find LCA of nodes n1 and n2 in Binary Tree Node *LCA(Node *n1, Node *n2) {    // Creata a map to store ancestors of n1    map <Node *, bool> ancestors;       // Insert n1 and all its ancestors in map    while (n1 != NULL)    {        ancestors[n1] = true;        n1 = n1->parent;    }       // Check if n2 or any of its ancestors is in    // map.    while (n2 != NULL)    {        if (ancestors.find(n2) != ancestors.end())            return n2;        n2 = n2->parent;    }       return NULL; }    // Driver method to test above functions int main(void) {     Node * root = NULL;        root = insert(root, 20);     root = insert(root, 8);     root = insert(root, 22);     root = insert(root, 4);     root = insert(root, 12);     root = insert(root, 10);     root = insert(root, 14);        Node *n1 = root->left->right->left;     Node *n2 = root->left;     Node *lca = LCA(n1, n2);        printf("LCA of %d and %d is %d ", n1->key, n2->key, lca->key);        return 0; }

Java

import java.util.HashMap; import java.util.Map;    // Java program to find lowest common ancestor using parent pointer // A tree node class Node  {     int key;     Node left, right, parent;        Node(int key)      {         this.key = key;         left = right = parent = null;     } }    class BinaryTree  {     Node root, n1, n2, lca;        /* A utility function to insert a new node with        given key in Binary Search Tree */     Node insert(Node node, int key)      {         /* If the tree is empty, return a new node */         if (node == null)             return new Node(key);            /* Otherwise, recur down the tree */         if (key < node.key)          {             node.left = insert(node.left, key);             node.left.parent = node;         }         else if (key > node.key)          {             node.right = insert(node.right, key);             node.right.parent = node;         }            /* return the (unchanged) node pointer */         return node;     }        // To find LCA of nodes n1 and n2 in Binary Tree     Node LCA(Node n1, Node n2)      {         // Creata a map to store ancestors of n1         Map<Node, Boolean> ancestors = new HashMap<Node, Boolean>();            // Insert n1 and all its ancestors in map         while (n1 != null)          {             ancestors.put(n1, Boolean.TRUE);             n1 = n1.parent;         }            // Check if n2 or any of its ancestors is in         // map.         while (n2 != null)          {             if (ancestors.containsKey(n2) != ancestors.isEmpty())                 return n2;             n2 = n2.parent;         }            return null;     }        // Driver method to test above functions     public static void main(String[] args)      {         BinaryTree tree = new BinaryTree();         tree.root = tree.insert(tree.root, 20);         tree.root = tree.insert(tree.root, 8);         tree.root = tree.insert(tree.root, 22);         tree.root = tree.insert(tree.root, 4);         tree.root = tree.insert(tree.root, 12);         tree.root = tree.insert(tree.root, 10);         tree.root = tree.insert(tree.root, 14);            tree.n1 = tree.root.left.right.left;         tree.n2 = tree.root.left;         tree.lca = tree.LCA(tree.n1, tree.n2);            System.out.println("LCA of " + tree.n1.key + " and " + tree.n2.key                 + " is " + tree.lca.key);     } }    // This code has been contributed by Mayank Jaiswal(mayank_24)


Output:

LCA of 10 and 8 is 8 

Note : The above implementation uses insert of Binary Search Tree to create a Binary Tree, but the function LCA is for any Binary Tree (not necessarily a Binary Search Tree).


Time Complexity : O(h) where h is height of Binary Tree if we use hash table to implement the solution (Note that the above solution uses map which takes O(Log h) time to insert and find). So the time complexity of above implementation is O(h Log h).

Auxiliary Space : O(h)

 

A O(h) time and O(1) Extra Space Solution:
The above solution requires extra space because we need to use a hash table to store visited ancestors. We can solve the problem in O(1) extra space using following fact : If both nodes are at same level and if we traverse up using parent pointers of both nodes, the first common node in the path to root is lca.
The idea is to find depths of given nodes and move up the deeper node pointer by the difference between depths. Once both nodes reach same level, traverse them up and return the first common node.

Thanks to Mysterious Mind for suggesting this approach.

C++

// C++ program to find lowest common ancestor using parent pointer #include <bits/stdc++.h> using namespace std;    // A Tree Node struct Node {     Node *left, *right, *parent;     int key; };    // A utility function to create a new BST node Node *newNode(int item) {     Node *temp = new Node;     temp->key = item;     temp->parent = temp->left = temp->right = NULL;     return temp; }    /* A utility function to insert a new node with given key in Binary Search Tree */ Node *insert(Node *node, int key) {     /* If the tree is empty, return a new node */     if (node == NULL) return newNode(key);        /* Otherwise, recur down the tree */     if (key < node->key)     {         node->left = insert(node->left, key);         node->left->parent = node;     }     else if (key > node->key)     {         node->right = insert(node->right, key);         node->right->parent = node;     }        /* return the (unchanged) node pointer */     return node; }    // A utility function to find depth of a node // (distance of it from root) int depth(Node *node) {     int d = -1;     while (node)     {         ++d;         node = node->parent;     }     return d; }    // To find LCA of nodes n1 and n2 in Binary Tree Node *LCA(Node *n1, Node *n2) {     // Find depths of two nodes and differences     int d1 = depth(n1), d2 = depth(n2);     int diff = d1 - d2;        // If n2 is deeper, swap n1 and n2     if (diff < 0)     {         Node * temp = n1;         n1 = n2;         n2 = temp;         diff = -diff;     }        // Move n1 up until it reaches the same level as n2     while (diff--)         n1 = n1->parent;        // Now n1 and n2 are at same levels     while (n1 && n2)     {         if (n1 == n2)             return n1;         n1 = n1->parent;         n2 = n2->parent;     }        return NULL; }    // Driver method to test above functions int main(void) {     Node * root = NULL;        root = insert(root, 20);     root = insert(root, 8);     root = insert(root, 22);     root = insert(root, 4);     root = insert(root, 12);     root = insert(root, 10);     root = insert(root, 14);        Node *n1 = root->left->right->left;     Node *n2 = root->right;        Node *lca = LCA(n1, n2);     printf("LCA of %d and %d is %d ", n1->key, n2->key, lca->key);        return 0; }

Java

import java.util.HashMap; import java.util.Map;    // Java program to find lowest common ancestor using parent pointer    // A tree node class Node  {     int key;     Node left, right, parent;        Node(int key)      {         this.key = key;         left = right = parent = null;     } }    class BinaryTree  {     Node root, n1, n2, lca;        /* A utility function to insert a new node with        given key in Binary Search Tree */     Node insert(Node node, int key)      {         /* If the tree is empty, return a new node */         if (node == null)             return new Node(key);            /* Otherwise, recur down the tree */         if (key < node.key)          {             node.left = insert(node.left, key);             node.left.parent = node;         }          else if (key > node.key)          {             node.right = insert(node.right, key);             node.right.parent = node;         }            /* return the (unchanged) node pointer */         return node;     }        // A utility function to find depth of a node     // (distance of it from root)     int depth(Node node)      {         int d = -1;         while (node != null)          {             ++d;             node = node.parent;         }         return d;     }        // To find LCA of nodes n1 and n2 in Binary Tree     Node LCA(Node n1, Node n2)      {         // Find depths of two nodes and differences         int d1 = depth(n1), d2 = depth(n2);         int diff = d1 - d2;            // If n2 is deeper, swap n1 and n2         if (diff < 0)          {             Node temp = n1;             n1 = n2;             n2 = temp;             diff = -diff;         }            // Move n1 up until it reaches the same level as n2         while (diff-- != 0)             n1 = n1.parent;            // Now n1 and n2 are at same levels         while (n1 != null && n2 != null)          {             if (n1 == n2)                 return n1;             n1 = n1.parent;             n2 = n2.parent;         }            return null;     }        // Driver method to test above functions     public static void main(String[] args)      {         BinaryTree tree = new BinaryTree();         tree.root = tree.insert(tree.root, 20);         tree.root = tree.insert(tree.root, 8);         tree.root = tree.insert(tree.root, 22);         tree.root = tree.insert(tree.root, 4);         tree.root = tree.insert(tree.root, 12);         tree.root = tree.insert(tree.root, 10);         tree.root = tree.insert(tree.root, 14);            tree.n1 = tree.root.left.right.left;         tree.n2 = tree.root.right;         tree.lca = tree.LCA(tree.n1, tree.n2);            System.out.println("LCA of " + tree.n1.key + " and " + tree.n2.key                 + " is " + tree.lca.key);     } }    // This code has been contributed by Mayank Jaiswal(mayank_24)

C#

// C# program to find lowest common  // ancestor using parent pointer  using System;    // A tree node  public class Node {     public int key;     public Node left, right, parent;        public Node(int key)     {         this.key = key;         left = right = parent = null;     } }    class GFG { public Node root, n1, n2, lca;    /* A utility function to insert a new  node with given key in Binary Search Tree */ public virtual Node insert(Node node, int key) {     /* If the tree is empty, return         a new node */     if (node == null)     {         return new Node(key);     }        /* Otherwise, recur down the tree */     if (key < node.key)     {         node.left = insert(node.left, key);         node.left.parent = node;     }     else if (key > node.key)     {         node.right = insert(node.right, key);         node.right.parent = node;     }        /* return the (unchanged) node pointer */     return node; }    // A utility function to find depth of a  // node (distance of it from root)  public virtual int depth(Node node) {     int d = -1;     while (node != null)     {         ++d;         node = node.parent;     }     return d; }    // To find LCA of nodes n1 and n2  // in Binary Tree  public virtual Node LCA(Node n1, Node n2) {     // Find depths of two nodes      // and differences      int d1 = depth(n1), d2 = depth(n2);     int diff = d1 - d2;        // If n2 is deeper, swap n1 and n2      if (diff < 0)     {         Node temp = n1;         n1 = n2;         n2 = temp;         diff = -diff;     }        // Move n1 up until it reaches      // the same level as n2      while (diff-- != 0)     {         n1 = n1.parent;     }        // Now n1 and n2 are at same levels      while (n1 != null && n2 != null)     {         if (n1 == n2)         {             return n1;         }         n1 = n1.parent;         n2 = n2.parent;     }        return null; }    // Driver Code public static void Main(string[] args) {     GFG tree = new GFG();     tree.root = tree.insert(tree.root, 20);     tree.root = tree.insert(tree.root, 8);     tree.root = tree.insert(tree.root, 22);     tree.root = tree.insert(tree.root, 4);     tree.root = tree.insert(tree.root, 12);     tree.root = tree.insert(tree.root, 10);     tree.root = tree.insert(tree.root, 14);        tree.n1 = tree.root.left.right.left;     tree.n2 = tree.root.right;     tree.lca = tree.LCA(tree.n1, tree.n2);        Console.WriteLine("LCA of " + tree.n1.key +                       " and " + tree.n2.key +                        " is " + tree.lca.key); } }    // This code is contributed by Shrikant13


Output :

LCA of 10 and 22 is 20 

You may like to see below articles as well :

Lowest Common Ancestor in a Binary Tree | Set 1

Lowest Common Ancestor in a Binary Search Tree.

Find LCA in Binary Tree using RMQ