CODE 17. Populating Next Right Pointers in Each Node

版权声明：本文为博主原创文章，转载请注明出处，谢谢!

版权声明：本文为博主原创文章，转载请注明出处：http://blog.jerkybible.com/2013/09/16/2013-09-16-CODE 17 Populating Next Right Pointers in Each Node/

访问原文「CODE 17. 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

public void connect(TreeLinkNode root) {
	// Start typing your Java solution below
	// DO NOT write main() function
	if (null == root) {
		return;
	}
	bfs(root);
}

private void bfs(TreeLinkNode root) {
	Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
	queue.offer(root);
	int queueNumber = 1;
	int layerNumber = 0;
	TreeLinkNode tempNode = null;
	while (!queue.isEmpty()) {
		tempNode = queue.poll();
		queueNumber--;
		if (null != tempNode.left) {
			queue.offer(tempNode.left);
			layerNumber++;
		}
		if (null != tempNode.right) {
			queue.offer(tempNode.right);
			layerNumber++;
		}
		if (queueNumber != 0) {
			tempNode.next = queue.peek();
		} else {
			tempNode.next = null;
		}

		if (queueNumber == 0 && layerNumber != 0) {
			queueNumber = layerNumber;
			layerNumber = 0;
		}
	}
}