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

这道题目用bfs 和 dfs 都可以做。更推荐bfs。如果用dfs ，则需要注意的 是 要先遍历右子树，再遍历左子树。因为对左子树的操作，用到了右子树的结果。

下面是我写的dfs，后面是答案的bfs

/**
* 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;
}
root.next = null;
find(root, root.right);
find(root, root.left);
}
private void find(TreeLinkNode parent, TreeLinkNode cur) {
if (cur == null) {
return;
}
if (parent.right != null && cur != parent.right) {
cur.next = parent.right;
} else {
TreeLinkNode temp = parent.next;
while (temp != null) {
if (temp.left != null) {
cur.next = temp.left;
break;
} else if (temp.right != null) {
cur.next = temp.right;
break;
} else{
temp = temp.next;
}
}
if (temp == null) {
cur.next = null;
}
}
if (cur.right != null) {
find(cur, cur.right);
}
if (cur.left != null) {
find(cur, cur.left);
}

}
}

public class Solution {
public void connect(TreeLinkNode root) {
if (root == null) {
return;
}

TreeLinkNode parent = root;
TreeLinkNode pre;
TreeLinkNode next;
while (parent != null) {
pre = null;
next = null;
while (parent != null) {
if (next == null){
next = (parent.left != null) ? parent.left: parent.right;
}

if (parent.left != null){
if (pre != null) {
pre.next = parent.left;
pre = pre.next;
} else {
pre = parent.left;
}
}

if (parent.right != null) {
if (pre != null) {
pre.next = parent.right;
pre = pre.next;
} else {
pre = parent.right;
}
}
parent = parent.next;
}
parent = next;
}
}
}