populating-next-right-pointers-in-each-node-ii

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1、populating-next-right-pointers-in-each-node-ii 来源：牛客网

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    7After calling your function, the tree should look like:1 -> NULL/  \2 -> 3 -> NULL/ \    \4-> 5 -> 7 -> NULL

2、思路：

代码1思路，采用队列，对于每一层的元素，遍历时，用一个临时的结点cur放置当前poll出来的结点，poll后，检查当前队列的对头是否是该层元素，若是用cur指向就可以了；
代码2思路，不使用队列（题目说是You may only use constant extra space）。首先不使用队列的情况下我们要遍历整个树，就得利用它每一层都有一条链，有链就可以通过一个结点遍历一层了。这里通过新建头部结点tmpLevelFirst（该结点本身的值和树没关系，只有tmpLevelFirst.next来指向每一层的头结点）来实现

3、代码1，用队列辅助实现：

public class BFSConnectTree {public void connect(TreeLinkNode root) {if(root == null)return;Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();queue.add(root);TreeLinkNode node = new TreeLinkNode(0);while(!queue.isEmpty()){int broad_size  = queue.size();//每一层 的节点数目for(int i = 0; i < broad_size; i++){node = queue.poll();if(i == broad_size - 1){//若是当前层的最后一个元素手动指向nullnode.next = null;}else{node.next = queue.peek();}if(node.left != null)queue.add(node.left);if(node.right != null)queue.add(node.right);}}}}

代码2，利用每层都是链的特点实现：

public class BFSConnectTreeWithoutSpace {public void connect(TreeLinkNode root) {while (root != null) {//tmpLevelFirst为新建立的每层第一个节点（为了方便后面遍历当前行节点）TreeLinkNode tmpLevelFirst = new TreeLinkNode(0);TreeLinkNode cur = tmpLevelFirst;while (root != null) {if (root.left != null) {cur.next = root.left;cur = cur.next;}if (root.right != null) {cur.next = root.right;cur = cur.next;}root = root.next;}root = tmpLevelFirst.next;}}
}