Populating Next Right Pointers in Each Node II

Given a binary tree

}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set toNULL.

Initially, all next pointers are set toNULL.

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

Example:

Given the following binary tree,

1
/  \
2    3
/ \    \
4   5    7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \    \
4-> 5 -> 7 -> NULL

Analysis

A very concise and clear level-order traversal (@davidtan1890):

public class Solution {
public void connect(TreeLinkNode root) {

while(root != null){
TreeLinkNode currentChild = tempChild;
while(root!=null){
if(root.left != null) {
currentChild.next = root.left;
currentChild = currentChild.next;
}
if(root.right != null) {
currentChild.next = root.right;
currentChild = currentChild.next;
}
root = root.next;
}
root = tempChild.next;
}
}
}

Iterative

Solution

Iterative - O(n) time, O(1) space

public class Solution {
public void connect(TreeLinkNode root) {
TreeLinkNode nextHead = new TreeLinkNode(0); // dummy node for next level
TreeLinkNode tail = null;
TreeLinkNode curr = null;
while (nextHead.next != null) {
while (curr != null) {
if (curr.left != null) {
tail.next = curr.left;
tail = tail.next;
}
if (curr.right != null) {
tail.next = curr.right;
tail = tail.next;
}
curr = curr.next;
}
}
}
}

Recursive

public class Solution {
public void connect(TreeLinkNode root) {
if (root == null)
return;
for (TreeLinkNode curr = root, prev = dummy; curr != null; curr = curr.next) {
if (curr.left != null){
prev.next = curr.left;
prev = prev.next;
}
if (curr.right != null) {
prev.next = curr.right;
prev = prev.next;
}
}
connect(dummy.next);
}
}

Reference

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/37828/O(1)-space-O(n)-complexity-Iterative-Solution

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/37828/O(1)-space-O(n)-complexity-Iterative-Solution/35897