Algorithm
Problem Nmae: 116. Populating Next Right Pointers in Each Node
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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
void connect(struct TreeLinkNode *root) {
struct TreeLinkNode *v, *h;
if (!root || !root->left) return;
v = root;
while (v->left) {
h = v;
do {
h->left->next = h->right;
if (h->next) {
h->right->next = h->next->left;
}
h = h->next;
} while (h);
v = v->left;
}
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) return;
deque < TreeLinkNode*>cur;
dequenext;
cur.push_back(root);
while(!cur.empty()){
TreeLinkNode* node = cur.front();
cur.pop_front();
node->next = cur.empty() ? NULL : cur.front();
if(node->left) next.push_back(node->left);
if(node->right) next.push_back(node->right);
if(cur.empty()) swap(cur, next);
}
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public Node connect(Node root) {
Node prev = root;
Node curr = null;
while (prev != null && prev.left != null) {
curr = prev;
while (curr != null) {
curr.left.next = curr.right;
if (curr.next != null) {
curr.right.next = curr.next.left;
}
curr = curr.next;
}
prev = prev.left;
}
return root;
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const connect = function(root) {
if (root == null) return null
const cur = [root]
while (cur.length) {
let len = cur.length
for (let i = 0; i < len; i++) {
let el = cur.shift()
if (i === len - 1) el.next = null
else el.next = cur[0]
if (el.left) cur.push(el.left)
if (el.right) cur.push(el.right)
}
}
return root
}
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def connect(self, root: "Node") -> "Node":
if root == None:
return root
q, prev = [(root, 1)], None
while q:
curr, pos = q.pop(0)
if prev != None and prev[1] == pos:
prev[0].next = curr
prev = [curr, pos]
if curr.left:
q.append((curr.left, pos + 1))
if curr.right:
q.append((curr.right, pos + 1))
return root
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0116_PopulatingNextRightPointersInEachNode
{
public TreeLinkNode Connect(TreeLinkNode root)
{
TreeLinkNode levelStart = root;
while (levelStart != null)
{
TreeLinkNode cur = levelStart;
while (cur != null)
{
if (cur.left != null) cur.left.next = cur.right;
if (cur.right != null && cur.next != null) cur.right.next = cur.next.left;
cur = cur.next;
}
levelStart = levelStart.left;
}
return root;
}
}
}
Input
Output