// 116. Populating Next Right Pointers in Each Node [**PREFETC BINARY TREE]
// Given the following perfect binary tree. after calling your function, the tree should look like:
// 1 1 -> NULL
// / \ / \
// 2 3 2 -> 3 -> NULL
// / \ / \ / \ / \
// 4 5 6 7 4->5->6->7 -> NULL
// solution 1: recursive
public static void PopulateNextRightPointer(Node2 root)
{
if (root == null)
{
return;
}
if (root.left != null)
{
root.left.next = root.right;
}
if (root.right != null && root.next != null)
{
root.right.next = root.next.left;
}
PopulateNextRightPointer(root.left);
PopulateNextRightPointer(root.right);
}
// solution 3: optimal O(1) space solution
public void PopulateNextRightPointer3(Node2 root)
{
if (root == null)
{
return;
}
Node2 start = root, node = null;
while (start.left != null)
{
node = start;
while (node != null)
{
node.left.next = node.right;
if (node.next != null)
{
node.right.next = node.next.left;
}
node = node.next;
}
start = start.left;
}
}
// solution 2[Java]: queue level-order traversal
public void PopulateNextRightPointer2(Node2 root)
{
if (root == null)
{
return;
}
Queue <Node2> q = new Queue <Node2>();
q.Enqueue(root);
while (q.Count > 0)
{
int size = q.Count;
for (int i = 0; i < size; i++)
{
Node2 node = q.Dequeue();
if (node.left != null)
{
node.left.next = node.right;
}
if (node.right != null && node.next != null)
{
node.right.next = node.next.left;
}
if (node.left != null)
{
q.Enqueue(node.left);
}
if (node.right != null)
{
q.Enqueue(node.right);
}
}
}
}
