117. 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

空间O(1)版本 http://www.2cto.com/kf/201311/259258.html

public static void connect(TreeLinkNode root) {
// 空节点就直接返回
if (root == null){
return;
}

// 找到与root同一行的next node
TreeLinkNode rootNext = root.next;
TreeLinkNode next = null;       // 下一个被连接的对象

// rootNext如果是null说明已经处理完这一层的所有node
// next不等于null说明找到了找到最左边的下一个被连接的对象
while (rootNext != null && next == null)
{
if (rootNext.left != null){ // 优先找左边
next = rootNext.left;
} else{
next = rootNext.right;
}
rootNext = rootNext.next;
}

if (root.left != null)
{
if (root.right != null){    //  内部相连
root.left.next = root.right;
}else{                      // 跨树相连
root.left.next = next;
}
}
if (root.right != null){        // 跨树相连
root.right.next = next;
}

connect(root.right);        // 要先让右边都先连起来
connect(root.left);
}

另外一种层次遍历的方法，最坏情况空间O(N/2)
public void connect(TreeLinkNode root)
{
if(root==null)
return ;

ArrayDeque<TreeLinkNode> deque=new ArrayDeque<>();
deque.add(root);
int level=1;
TreeLinkNode pre=null;
boolean levelchang=true;

while(!deque.isEmpty())
{
TreeLinkNode t=deque.poll();

if(t.left!=null)
deque.add(t.left);

if(t.right!=null)
deque.add(t.right);

if(!levelchang)
pre.next=t;

level--;
if(level==0)
{
levelchang=true;
level=deque.size();
}
else {
levelchang=false;
}

pre=t;
}

}