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Sum of Nodes with Even-Valued Grandparent in C++
Suppose we have a binary tree, we have to find the sum of values of nodes with even-valued grandparent. (A grandparent of a node is the parent of its parent, if it exists.). If there are no such nodes with an even-valued grandparent, then return 0. So if the tree is like −
The output will be 18. The red nodes are nodes with even-value grandparent, while the blue nodes are the even valued grandparents.
To solve this, we will follow these steps −
- Define a map called parent
- Define a method called solve(), this will take node and par
- if node is null, then return
- if par is not null and par is present in parent and parent[par] is not 0 and value of parent[par] is even, then
- res := res + value of node
- parent[node] := par
- solve(left of node, node)
- solve(right of node, node)
- From the main method, set res := 0, call solve(root, Null), then return res
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class TreeNode{
public:
int val;
TreeNode *left, *right;
TreeNode(int data){
val = data;
left = NULL;
right = NULL;
}
};
void insert(TreeNode **root, int val){
queue<TreeNode*> q;
q.push(*root);
while(q.size()){
TreeNode *temp = q.front();
q.pop();
if(!temp->left){
if(val != NULL)
temp->left = new TreeNode(val);
else
temp->left = new TreeNode(0);
return;
}
else{
q.push(temp->left);
}
if(!temp->right){
if(val != NULL)
temp->right = new TreeNode(val);
else
temp->right = new TreeNode(0);
return;
}
else{
q.push(temp->right);
}
}
}
TreeNode *make_tree(vector<int> v){
TreeNode *root = new TreeNode(v[0]);
for(int i = 1; i<v.size(); i++){
insert(&root, v[i]);
}
return root;
}
class Solution {
public:
int res;
map <TreeNode*, TreeNode*> parent;
void solve(TreeNode* node, TreeNode* par = NULL){
if(!node)return;
if(par && parent.count(par) && parent[par] && parent[par]->val % 2 == 0){
res += node->val;
}
parent[node] = par;
solve(node->left, node);
solve(node->right, node);
}
int sumEvenGrandparent(TreeNode* root) {
res = 0;
parent.clear();
solve(root);
return res;
}
};
main(){
vector<int> v = {6,7,8,2,7,1,3,9,NULL,1,4,NULL,NULL,NULL,5};
TreeNode *root = make_tree(v);
Solution ob;
cout << (ob.sumEvenGrandparent(root));
}Input
[6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
Output
18
Updated on: 30-Apr-2020
89 Views
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