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node_ancestors_in_root_path.cpp
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node_ancestors_in_root_path.cpp
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/**
* Given a binary Tree
* 1
* / \
* 2 3
* / \
* 4 5
* \
* 6
*
* Ancestor of node 6 in root path would be 4, 2, and 1.
*
*/
#include <iostream>
#include <vector>
struct Node {
int data;
Node * left;
Node * right;
Node( int d ) : data{ d }, left{ nullptr }, right{ nullptr } { }
};
bool get_ancestors_util( Node * root, int key, std::vector<int> & ancestors ) {
if ( root == nullptr ) {
return false;
}
if ( root->data == key ) {
return true;
}
if ( get_ancestors_util(root->left, key, ancestors) ||
get_ancestors_util(root->right, key, ancestors) ) {
ancestors.push_back(root->data);
return true;
}
return false;
}
std::vector<int> get_ancestors( Node * root, int key ) {
std::vector<int> ancestors;
get_ancestors_util(root, key, ancestors);
return ancestors;
}
void printAncestors( Node * root ) {
std::vector<int> vec;
for ( int i = 1; i <= 6; ++i ) {
std::cout << "Ancestor of node with key " << i << " are:";
vec = get_ancestors( root, i );
for ( auto v : vec ) {
std::cout << v << " ";
}
std::cout << std::endl;
vec.clear();
}
}
int main() {
Node * root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
root->left->left->right = new Node(6);
printAncestors(root);
return 0;
}