Given a binary tree (not a binary search tree) and any number of Key Nodes, the task is to find the least common ancestor of all the key Nodes.
Following is the definition of LCA from Wikipedia:
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).
The LCA of any number of nodes in T is the shared common ancestor of the nodes that is located farthest from the root.
Example: In the figure above:
LCA of nodes 12, 14, 15 is node 3 LCA of nodes 3, 14, 15 is node 3 LCA of nodes 6, 7, 15 is node 3 LCA of nodes 5, 13, 14, 15 is node 1 LCA of nodes 6, 12 is node 6
Approach:
Following is the simple approach for Least Common Ancestor for any number of nodes.
- For every node calculate the matching number of nodes at that node and its sub-tree.
- If root is also a matching node.
matchingNodes = matchingNodes in left sub-tree + matchingNodes in right sub-tree + 1
- If root is not a matching node.
matchingNodes = matchingNodes in left sub-tree + matchingNodes in right-subtree
- If matching Nodes count at any node is equal to number of keys then add that node into the Ancestors list.
- The First node in the Ancestors List is the Least Common Ancestor of all the given keys.
Below is the implementation of above approach.
C++
// C++ implementation to find// Ancestors of any number of nodes#include <bits/stdc++.h>using namespace std;// Tree Classclass TreeNode { public: int data; TreeNode *left, *right; TreeNode(int value) { this->data = value; this->left = NULL; this->right = NULL; }};int getKeysCount(TreeNode *root, vector<int> &keyNodes, int matchingNodes, vector<TreeNode *> &ancestors){ // Base Case. When root is Null if (root == NULL) { return 0; } // Search for left and right subtree // for matching child Key Node. matchingNodes += getKeysCount(root->left, keyNodes, matchingNodes, ancestors) + getKeysCount(root->right, keyNodes, matchingNodes, ancestors); // Condition to check if Root Node // is also in Key Node if (find(keyNodes.begin(), keyNodes.end(), root->data) != keyNodes.end()) { matchingNodes++; } // Condition when matching Nodes is // equal to the Key Nodes found if (matchingNodes == keyNodes.size()) { ancestors.push_back(root); } return matchingNodes;}// Function to find Least Common// Ancestors of N number of nodesTreeNode *lcaOfNodes(TreeNode *root, vector<int> &keyNodes){ // Create a new list for // capturing all the ancestors // of the given nodes vector<TreeNode *> ancestors; // Initially there is No Matching Nodes int matchingNodes = 0; getKeysCount(root, keyNodes, matchingNodes, ancestors); // First Node in the Ancestors list // is the Least Common Ancestor of // Given keyNodes return ancestors[0];}// Driver Codeint main() { // Creation of Tree TreeNode *root = new TreeNode(1); root->left = new TreeNode(2); root->right = new TreeNode(3); root->left->left = new TreeNode(4); root->left->right = new TreeNode(5); root->right->left = new TreeNode(6); root->right->right = new TreeNode(7); root->left->left->left = new TreeNode(8); root->left->left->right = new TreeNode(9); root->left->right->left = new TreeNode(10); root->left->right->right = new TreeNode(11); root->right->left->left = new TreeNode(12); root->right->left->right = new TreeNode(13); root->right->right->left = new TreeNode(14); root->right->right->right = new TreeNode(15); // Key Nodes for LCA vector<int> keyNodes; keyNodes.push_back(12); keyNodes.push_back(14); keyNodes.push_back(15); cout << lcaOfNodes(root, keyNodes)->data << endl; return 0;}// This code is contributed by sanjeev2552 |
Java
// Java implementation to find// Ancestors of any number of nodesimport java.util.ArrayList;// Tree Classclass TreeNode { int data; TreeNode left; TreeNode right; public TreeNode(int value) { this.data = value; left = right = null; }}public class LCAofAnyNumberOfNodes { // Function to find Least Common // Ancestors of N number of nodes public static TreeNode lcaOfNodes( TreeNode root, ArrayList<Integer> keyNodes) { // Create a new list for // capturing all the ancestors // of the given nodes ArrayList<TreeNode> ancestors = new ArrayList<TreeNode>(); // Initially there is No Matching Nodes int matchingNodes = 0; getKeysCount(root, keyNodes, matchingNodes, ancestors); // First Node in the Ancestors list // is the Least Common Ancestor of // Given keyNodes return ancestors.get(0); } private static int getKeysCount( TreeNode root, ArrayList<Integer> keyNodes, int matchingNodes, ArrayList<TreeNode> ancestors) { // Base Case. When root is Null if (root == null) return 0; // Search for left and right subtree // for matching child Key Node. matchingNodes += getKeysCount(root.left, keyNodes, matchingNodes, ancestors) + getKeysCount(root.right, keyNodes, matchingNodes, ancestors); // Condition to check if Root Node // is also in Key Node if (keyNodes.contains(root.data)){ matchingNodes++; } // Condition when matching Nodes is // equal to the Key Nodes found if (matchingNodes == keyNodes.size()) ancestors.add(root); return matchingNodes; } // Driver Code public static void main(String[] args) { // Creation of Tree TreeNode root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); root.left.right = new TreeNode(5); root.right.left = new TreeNode(6); root.right.right = new TreeNode(7); root.left.left.left = new TreeNode(8); root.left.left.right = new TreeNode(9); root.left.right.left = new TreeNode(10); root.left.right.right = new TreeNode(11); root.right.left.left = new TreeNode(12); root.right.left.right = new TreeNode(13); root.right.right.left = new TreeNode(14); root.right.right.right = new TreeNode(15); // Key Nodes for LCA ArrayList<Integer> keyNodes = new ArrayList<Integer>(); keyNodes.add(12); keyNodes.add(14); keyNodes.add(15); System.out.println( lcaOfNodes(root, keyNodes).data ); }} |
Python3
# Python3 implementation to find# Ancestors of any number of nodes# Tree Classclass TreeNode: def __init__(self, value): self.data = value self.left = None self.right = None# Create a new list for# capturing all the ancestors# of the given nodesancestors = [] # Function to find Least Common# Ancestors of N number of nodesdef lcaOfNodes(root, keyNodes): # Initially there is No Matching Nodes matchingNodes = 0 getKeysCount(root, keyNodes, matchingNodes) # First Node in the Ancestors list # is the Least Common Ancestor of # Given keyNodes ancestors[0].data-=1 return ancestors[0] def getKeysCount(root, keyNodes, matchingNodes): # Base Case. When root is Null if root == None: return 0 # Search for left and right subtree # for matching child Key Node. matchingNodes += getKeysCount(root.left, keyNodes, matchingNodes) + getKeysCount(root.right, keyNodes, matchingNodes) # Condition to check if Root Node # is also in Key Node if keyNodes: matchingNodes+=1 # Condition when matching Nodes is # equal to the Key Nodes found if matchingNodes == len(keyNodes): ancestors.append(root) return matchingNodes# Creation of Treeroot = TreeNode(1)root.left = TreeNode(2)root.right = TreeNode(3)root.left.left = TreeNode(4)root.left.right = TreeNode(5)root.right.left = TreeNode(6)root.right.right = TreeNode(7)root.left.left.left = TreeNode(8)root.left.left.right = TreeNode(9)root.left.right.left = TreeNode(10)root.left.right.right = TreeNode(11)root.right.left.left = TreeNode(12)root.right.left.right = TreeNode(13)root.right.right.left = TreeNode(14)root.right.right.right = TreeNode(15) # Key Nodes for LCAkeyNodes = []keyNodes.append(12)keyNodes.append(14)keyNodes.append(15)tmp = lcaOfNodes(root, keyNodes)print(tmp.data)# This code is contributed by suresh07. |
C#
// C# implementation to find// Ancestors of any number of nodesusing System;using System.Collections.Generic;// Tree Classclass TreeNode { public int data; public TreeNode left; public TreeNode right; public TreeNode(int value) { this.data = value; left = right = null; }} public class LCAofAnyNumberOfNodes { // Function to find Least Common // Ancestors of N number of nodes static TreeNode lcaOfNodes( TreeNode root, List<int> keyNodes) { // Create a new list for // capturing all the ancestors // of the given nodes List<TreeNode> ancestors = new List<TreeNode>(); // Initially there is No Matching Nodes int matchingNodes = 0; getKeysCount(root, keyNodes, matchingNodes, ancestors); // First Node in the Ancestors list // is the Least Common Ancestor of // Given keyNodes return ancestors[0]; } private static int getKeysCount( TreeNode root, List<int> keyNodes, int matchingNodes, List<TreeNode> ancestors) { // Base Case. When root is Null if (root == null) return 0; // Search for left and right subtree // for matching child Key Node. matchingNodes += getKeysCount(root.left, keyNodes, matchingNodes, ancestors) + getKeysCount(root.right, keyNodes, matchingNodes, ancestors); // Condition to check if Root Node // is also in Key Node if (keyNodes.Contains(root.data)){ matchingNodes++; } // Condition when matching Nodes is // equal to the Key Nodes found if (matchingNodes == keyNodes.Count) ancestors.Add(root); return matchingNodes; } // Driver Code public static void Main(String[] args) { // Creation of Tree TreeNode root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); root.left.right = new TreeNode(5); root.right.left = new TreeNode(6); root.right.right = new TreeNode(7); root.left.left.left = new TreeNode(8); root.left.left.right = new TreeNode(9); root.left.right.left = new TreeNode(10); root.left.right.right = new TreeNode(11); root.right.left.left = new TreeNode(12); root.right.left.right = new TreeNode(13); root.right.right.left = new TreeNode(14); root.right.right.right = new TreeNode(15); // Key Nodes for LCA List<int> keyNodes = new List<int>(); keyNodes.Add(12); keyNodes.Add(14); keyNodes.Add(15); Console.WriteLine( lcaOfNodes(root, keyNodes).data ); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation to find// Ancestors of any number of nodes// Tree Classclass TreeNode { constructor(value) { this.data = value; this.left = null; this.right = null; }} // Create a new list for // capturing all the ancestors// of the given nodesvar ancestors = [];// Function to find Least Common // Ancestors of N number of nodesfunction lcaOfNodes( root, keyNodes){ // Initially there is No Matching Nodes var matchingNodes = 0; getKeysCount(root, keyNodes, matchingNodes); // First Node in the Ancestors list // is the Least Common Ancestor of // Given keyNodes return ancestors[0];}function getKeysCount( root, keyNodes, matchingNodes){ // Base Case. When root is Null if (root == null) return 0; // Search for left and right subtree // for matching child Key Node. matchingNodes += getKeysCount(root.left, keyNodes, matchingNodes) + getKeysCount(root.right, keyNodes, matchingNodes); // Condition to check if Root Node // is also in Key Node if (keyNodes.includes(root.data)){ matchingNodes++; } // Condition when matching Nodes is // equal to the Key Nodes found if (matchingNodes == keyNodes.length) ancestors.push(root); return matchingNodes;} // Driver Code// Creation of Treevar root = new TreeNode(1);root.left = new TreeNode(2);root.right = new TreeNode(3);root.left.left = new TreeNode(4);root.left.right = new TreeNode(5);root.right.left = new TreeNode(6);root.right.right = new TreeNode(7);root.left.left.left = new TreeNode(8);root.left.left.right = new TreeNode(9);root.left.right.left = new TreeNode(10);root.left.right.right = new TreeNode(11);root.right.left.left = new TreeNode(12);root.right.left.right = new TreeNode(13);root.right.right.left = new TreeNode(14);root.right.right.right = new TreeNode(15); // Key Nodes for LCAvar keyNodes = [];keyNodes.push(12);keyNodes.push(14);keyNodes.push(15);var tmp = lcaOfNodes(root, keyNodes);document.write(tmp.data);</script> |
Output:
3
The time complexity of the lcaOfNodes() function is O(N), where N is the number of nodes in the tree, because the function traverses the entire tree once. The space complexity of the function is O(H + K), where H is the height of the tree and K is the number of key nodes, because the function maintains a list of ancestors which can have a maximum of H nodes, and it also stores the key nodes which can have a maximum of K nodes.
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