Leetcode Populating Next Right Pointers in Each Node II problem solution

In this Leetcode Populating Next Right Pointers in Each Node II problem solution we have Given a binary tree

struct Node {

int val;

Node *left;

Node *right;

Node *next;

}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. Initially, all next pointers are set to NULL.

Problem solution in Python.

```class Solution:
def connect(self, root: 'Node') -> 'Node':
ans=[]
def dfs(node,level):
if not(node):
return
if level==len(ans):
ans.append([])
if not(ans[level]):
ans[level].append(node)
else:
ans[level][-1].next=node
ans[level].append(node)
dfs(node.left,level+1)
dfs(node.right,level+1)
dfs(root,0)
return root
```

Problem solution in Java.

```public class Solution {
connect(root, null);
}

if (root == null) return;
root.next = right;
connect(root.right, nextRight);
connect(root.left, root.right != null? root.right : nextRight);
}

if (root == null) return null;
TreeLinkNode nextRight = root.left != null? root.left : root.right;
return nextRight != null? nextRight : getNextRight(root.next);
}
}
```

Problem solution in C++.

```class Solution {
public:
Node* connect(Node* root) {
if (!root) return root;
queue<Node*>q;
q.push(root);
while (q.size()){
int sz = q.size();
for (int i = 0; i < sz; i ++){
auto p = q.front(); q.pop();
if (i != sz - 1) p->next = q.front();
else p->next = NULL;
if (p->left) q.push(p->left);
if (p->right) q.push(p->right);
}
}
return root;
}
};
```

Problem solution in C.

```void dfs_fun(struct TreeLinkNode* node,struct TreeLinkNode* next) {
if(NULL==node) return;
node->next=next;
while(NULL!=next){
if(NULL!=next->left){
beyond_r=next->left;
break;
}

if(NULL!=next->right){
beyond_r=next->right;
break;
}

next=next->next;
}

dfs_fun(node->right,beyond_r);
dfs_fun(node->left,node->right?node->right:beyond_r);
}