116. 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

Solution 1

public static void connect2(TreeLinkNode root) {
if(root == null){
return;
}
TreeLinkNode pre = root;
TreeLinkNode cur = null;
while(pre.left != null){
cur = pre;
while(cur != null){
cur.left.next = cur.right;
if(cur.next != null){
cur.right.next = cur.next.left;
}
cur = cur.next;
}
pre = pre.left;
}
}

//same idea
public void connect22(TreeLinkNode root) {
while(root != null && root.left != null) {
TreeLinkNode cur = root;
while(cur != null) {
cur.left.next = cur.right;
cur.right.next = cur.next == null ? null : cur.next.left;
cur = cur.next;
}
root = root.left;
}
}

Solution 2 Recursive

public void connect3(TreeLinkNode root) {
if(root == null || root.left == null) return;
connectNodes(root.left, root.right);
}

public void connectNodes(TreeLinkNode node1, TreeLinkNode node2) {
node1.next = node2;
if(node1.left != null) {
connectNodes(node1.left, node1.right);
connectNodes(node1.right, node2.left);
connectNodes(node2.left, node2.right);
}
}