Populating Next Right Pointers in Each Node I, II

Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *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

.

Note:
You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

breadth-first-search

1.use a queue to breadth traverse the whole tree.

2. If i == size - 1, the next points to null, else the next points to queue.peek(). This method needs an extra space O(n) to store the
elements in the tree.

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if (root == null)
return;

Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
queue.offer(root);
while( !queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeLinkNode node = queue.poll();
if (node.left != null)
queue.offer(node.left);

if (node.right != null)
queue.offer(node.right);

if (i == size - 1)
node.next = null;

else
node.next = queue.peek();
}
}
}
}

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
TreeLinkNode levelStart = root;
TreeLinkNode cur = root;

while (levelStart != null) {
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;
}
}
}

Populating Next Right Pointers in Each Node II

 

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:
You may only use constant extra space.

For 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

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if (root == null)
return;

TreeLinkNode p = root.next;

while(p != null) {
if (p.left != null) {
p = p.left;
break;
}

if (p.right != null) {
p = p.right;
break;
}

p = p.next;
}

if (root.right != null) {
root.right.next = p;
}
if (root.left != null) {
root.left.next = (root.right != null ? root.right:p);
}

connect(root.right);
connect(root.left);
}
}