Populating Next Right Pointers in Each Node
Problem Statement :
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
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.
Example 1:
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
Example 2:
Input: root = []
Output: []
Constraints:
The number of nodes in the tree is in the range [0, 212 - 1].
-1000 <= Node.val <= 1000
Follow-up:
You may only use constant extra space.
The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.
Solution :
Solution in C :
/**
* Definition for a Node.
* struct Node {
* int val;
* struct Node *left;
* struct Node *right;
* struct Node *next;
* };
*/
void connect_left_to_right(struct Node* left, struct Node* right)
{
if (left && right) {
left->next = right;
connect_left_to_right(left->right, right->left);
}
}
struct Node* connect(struct Node* root)
{
if (root) {
connect_left_to_right(root->left, root->right);
connect(root->left);
connect(root->right);
}
return root;
}
Solution in C++ :
class Solution {
public:
Node* connect(Node* root) {
if(root == nullptr)
return root;
queue<Node*> queue;
queue.push(root);
while(queue.size() > 0)
{
deque<Node*> dq;
int length = queue.size();
for(int i = 0;i<length;i++)
{
Node* curr = queue.front();
queue.pop();
dq.push_back(curr);
if(curr->left!=nullptr)
queue.push(curr->left);
if(curr->right!=nullptr)
queue.push(curr->right);
}
while(dq.size() > 1)
{
Node* popped = dq.front();
dq.pop_front();
popped->next = dq.front();
}
Node* popped = dq.front();
dq.pop_front();
popped->next = nullptr;
}
return root;
}
};
Solution in Java :
class Solution {
public Node connect(Node root) {
if(root == null)
return root;
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while(queue.size() > 0)
{
Deque<Node> dq = new ArrayDeque<>();
int length = queue.size();
for(int i = 0;i<length;i++)
{
Node curr = queue.poll();
dq.addLast(curr);
if(curr.left!=null)
queue.add(curr.left);
if(curr.right!=null)
queue.add(curr.right);
}
while(dq.size() > 1)
{
Node popped = dq.removeFirst();
popped.next = dq.getFirst();
}
Node popped = dq.removeFirst();
popped.next = null;
}
return root;
}
}
Solution in Python :
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
queue = []
queue.append(root)
while queue:
dq = collections.deque()
length = len(queue)
for i in range(length):
curr = queue.pop(0)
dq.append(curr)
if curr.left:
queue.append(curr.left)
if curr.right:
queue.append(curr.right)
while len(dq) > 1:
popped = dq.popleft()
popped.next = dq[0]
popped = dq.popleft()
popped.next = None
return root
View More Similar Problems
Subsequence Weighting
A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =
View Solution →Kindergarten Adventures
Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti
View Solution →Mr. X and His Shots
A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M
View Solution →Jim and the Skyscrapers
Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space
View Solution →Palindromic Subsets
Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t
View Solution →Counting On a Tree
Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n
View Solution →