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This approach uses a breadth-first search (BFS) strategy by employing a queue to keep track of nodes at the current level and connect their children nodes from left to right. After processing all nodes on a level, we move to the children (if any), ensuring the next pointers are correctly set.
Time Complexity: O(n), where n is the number of nodes in the tree since each node is enqueued and dequeued once.
Space Complexity: O(n) for the queue in the worst case where the last level of the tree has a maximum number of nodes.
1function Node(val, left, right, next) {
2 this.val = val;
3 this.left = left;
4 this.right = right;
5 this.next = next;
6};
7
8var connect = function(root) {
9 if (!root) return null;
10 let queue = [];
11 queue.push(root);
12 while (queue.length) {
13 let size = queue.length;
14 let prev = null;
15 for (let i = 0; i < size; i++) {
16 let node = queue.shift();
17 if (prev) prev.next = node;
18 prev = node;
19 if (node.left) queue.push(node.left);
20 if (node.right) queue.push(node.right);
21 }
22 }
23 return root;
24};This JavaScript function organizes nodes in a level-order sequence using an array as a queue substitute. By adjusting the next pointers, it ensures each node links to the next right node correctly.
This approach leverages a two-pointer or head-tail strategy to eliminate the need for extra storage space beyond two pointers. The idea is to work with two nested loops; an outer loop goes level by level, and an inner loop connects nodes within the same level by their next pointers.
Time Complexity: O(n) since each node is processed once.
Space Complexity: O(1), only utilizing a few pointers.
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
}
public class Solution {
public void Connect(Node root) {
Node current = root;
Node dummy = new Node(0, null, null, null);
while (current != null) {
Node tail = dummy;
dummy.next = null;
while (current != null) {
if (current.left != null) {
tail.next = current.left;
tail = tail.next;
}
if (current.right != null) {
tail.next = current.right;
tail = tail.next;
}
current = current.next;
}
current = dummy.next;
}
}
}
The C# version employs a nested structure and tracks the position of the next level start through a dummy node, thereby setting next pointers efficiently without additional space use.