[leetcode] Populating Next Right Pointers in Each Node

From : https://leetcode.com/problems/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

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
*  int val;
*  TreeLinkNode *left, *right, *next;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
TreeLinkNode *cur=root, *up, *dp, *t;
while(cur) {
up = cur;
dp = t = cur->left;
while(dp) {
dp->next = up->right;
if(!up->next) break;
up = up->next;
dp = dp->next;
dp->next = up->left;
dp = dp->next;
}
cur = t;
}
}
};

/**
* 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 up = root;
while(up != null) {
TreeLinkNode dp = up.left;
TreeLinkNode newUp = dp;
while(dp != null) {
dp = dp.next = up.right;
if(up.next != null) {
up = up.next;
dp = dp.next = up.left;
} else {
dp = null;
}
}
up = newUp;
}
}
}