Iterative method to find ancestors of a given binary tree
Given a binary tree, print all the ancestors of a particular key existing in the tree without using recursion.
Here we will be discussing the implementation for the above problem.
Examples:
Input :
1
/ \
2 7
/ \ / \
3 5 8 9
/ \ /
4 6 10
Key = 6
Output : 5 2 1
Ancestors of 6 are 5, 2 and 1.
The idea is to use iterative postorder traversal of given binary tree.
C++
// C++ program to print all ancestors of a given key #include <bits/stdc++.h> using namespace std; // Structure for a tree node struct Node { int data; struct Node* left, *right; }; // A utility function to create a new tree node struct Node* newNode(int data) { struct Node* node = (struct Node*)malloc(sizeof(struct Node)); node->data = data; node->left = node->right = NULL; return node; } // Iterative Function to print all ancestors of a // given key void printAncestors(struct Node* root, int key) { if (root == NULL) return; // Create a stack to hold ancestors stack<struct Node*> st; // Traverse the complete tree in postorder way till // we find the key while (1) { // Traverse the left side. While traversing, push // the nodes into the stack so that their right // subtrees can be traversed later while (root && root->data != key) { st.push(root); // push current node root = root->left; // move to next node } // If the node whose ancestors are to be printed // is found, then break the while loop. if (root && root->data == key) break; // Check if right sub-tree exists for the node at top // If not then pop that node because we don't need // this node any more. if (st.top()->right == NULL) { root = st.top(); st.pop(); // If the popped node is right child of top, // then remove the top as well. Left child of // the top must have processed before. while (!st.empty() && st.top()->right == root) { root = st.top(); st.pop(); } } // if stack is not empty then simply set the root // as right child of top and start traversing right // sub-tree. root = st.empty() ? NULL : st.top()->right; } // If stack is not empty, print contents of stack // Here assumption is that the key is there in tree while (!st.empty()) { cout << st.top()->data << " "; st.pop(); } } // Driver program to test above functions int main() { // Let us construct a binary tree struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(7); root->left->left = newNode(3); root->left->right = newNode(5); root->right->left = newNode(8); root->right->right = newNode(9); root->left->left->left = newNode(4); root->left->right->right = newNode(6); root->right->right->left = newNode(10); int key = 6; printAncestors(root, key); return 0; } |
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Java
// Java program to print all // ancestors of a given key import java.util.*; class GfG { // Structure for a tree node static class Node { int data; Node left, right; } // A utility function to // create a new tree node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return node; } // Iterative Function to print // all ancestors of a given key static void printAncestors(Node root, int key) { if (root == null) return; // Create a stack to hold ancestors Stack<Node> st = new Stack<Node> (); // Traverse the complete tree in // postorder way till we find the key while (1 == 1) { // Traverse the left side. While // traversing, push the nodes into // the stack so that their right // subtrees can be traversed later while (root != null && root.data != key) { st.push(root); // push current node root = root.left; // move to next node } // If the node whose ancestors // are to be printed is found, // then break the while loop. if (root != null && root.data == key) break; // Check if right sub-tree exists // for the node at top If not then // pop that node because we don't // need this node any more. if (st.peek().right == null) { root = st.peek(); st.pop(); // If the popped node is right child of top, // then remove the top as well. Left child of // the top must have processed before. while (!st.isEmpty() && st.peek().right == root) { root = st.peek(); st.pop(); } } // if stack is not empty then simply // set the root as right child of // top and start traversing right // sub-tree. root = st.isEmpty() ? null : st.peek().right; } // If stack is not empty, print contents of stack // Here assumption is that the key is there in tree while (!st.isEmpty()) { System.out.print(st.peek().data + " "); st.pop(); } } // Driver code public static void main(String[] args) { // Let us construct a binary tree Node root = newNode(1); root.left = newNode(2); root.right = newNode(7); root.left.left = newNode(3); root.left.right = newNode(5); root.right.left = newNode(8); root.right.right = newNode(9); root.left.left.left = newNode(4); root.left.right.right = newNode(6); root.right.right.left = newNode(10); int key = 6; printAncestors(root, key); } } // This code is contributed // by prerna saini |
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Python3
# Python program to print all ancestors of a given key # A class to create a new tree node class newNode: def __init__(self, data): self.data = data self.left = self.right = None # Iterative Function to print all ancestors of a # given key def printAncestors(root, key): if (root == None): return # Create a stack to hold ancestors st = [] # Traverse the complete tree in postorder way till # we find the key while (1): # Traverse the left side. While traversing, push # the nodes into the stack so that their right # subtrees can be traversed later while (root and root.data != key): st.append(root) # push current node root = root.left # move to next node # If the node whose ancestors are to be printed # is found, then break the while loop. if (root and root.data == key): break # Check if right sub-tree exists for the node at top # If not then pop that node because we don't need # this node any more. if (st[-1].right == None): root = st[-1] st.pop() # If the popped node is right child of top, # then remove the top as well. Left child of # the top must have processed before. while (len(st) != 0 and st[-1].right == root): root = st[-1] st.pop() # if stack is not empty then simply set the root # as right child of top and start traversing right # sub-tree. root = None if len(st) == 0 else st[-1].right # If stack is not empty, print contents of stack # Here assumption is that the key is there in tree while (len(st) != 0): print(st[-1].data,end = " ") st.pop() # Driver code if __name__ == '__main__': # Let us construct a binary tree root = newNode(1) root.left = newNode(2) root.right = newNode(7) root.left.left = newNode(3) root.left.right = newNode(5) root.right.left = newNode(8) root.right.right = newNode(9) root.left.left.left = newNode(4) root.left.right.right = newNode(6) root.right.right.left = newNode(10) key = 6 printAncestors(root, key) # This code is contributed by PranchalK. |
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C#
// C# program to print all // ancestors of a given key using System; using System.Collections.Generic; class GfG { // Structure for a tree node public class Node { public int data; public Node left, right; } // A utility function to // create a new tree node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return node; } // Iterative Function to print // all ancestors of a given key static void printAncestors(Node root, int key) { if (root == null) return; // Create a stack to hold ancestors Stack<Node> st = new Stack<Node> (); // Traverse the complete tree in // postorder way till we find the key while (1 == 1) { // Traverse the left side. While // traversing, push the nodes into // the stack so that their right // subtrees can be traversed later while (root != null && root.data != key) { st.Push(root); // push current node root = root.left; // move to next node } // If the node whose ancestors // are to be printed is found, // then break the while loop. if (root != null && root.data == key) break; // Check if right sub-tree exists // for the node at top If not then // pop that node because we don't // need this node any more. if (st.Peek().right == null) { root = st.Peek(); st.Pop(); // If the popped node is right child of top, // then remove the top as well. Left child of // the top must have processed before. while (st.Count != 0 && st.Peek().right == root) { root = st.Peek(); st.Pop(); } } // if stack is not empty then simply // set the root as right child of // top and start traversing right // sub-tree. root = st.Count == 0 ? null : st.Peek().right; } // If stack is not empty, print contents of stack // Here assumption is that the key is there in tree while (st.Count != 0) { Console.Write(st.Peek().data + " "); st.Pop(); } } // Driver code public static void Main(String[] args) { // Let us construct a binary tree Node root = newNode(1); root.left = newNode(2); root.right = newNode(7); root.left.left = newNode(3); root.left.right = newNode(5); root.right.left = newNode(8); root.right.right = newNode(9); root.left.left.left = newNode(4); root.left.right.right = newNode(6); root.right.right.left = newNode(10); int key = 6; printAncestors(root, key); } } // This code has been contributed by 29AjayKumar |
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Output:
5 2 1
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