Populating Next Right Pointers in Each Node @LeetCode

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

	package leetcode.tree;
	import java.util.LinkedList;
	import java.util.Queue;
	/**
	* Solution: BFS. Only difference is connect nodes.
	* if (i == size - 1) --> node.next = null. if not , node.next = queue.peek();
	*
	* Don't use queue.isEmpty as if condition, at that time, queue is not empty. Next level nodes already stores in.
	*
	* @author jeffwan
	* @date Mar 11, 2014
	*/
	public class ConnectRightPointers {
	public void connect(TreeLinkNode root) {
	if (root == null) {
	return;
	}
	Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
	queue.offer(root);
	while (!queue.isEmpty()) {
	int size = queue.size();
	for (int i = 0; i < size; i++) {
	TreeLinkNode node = queue.poll();
	if (i == size - 1) {
	node.next = null;
	} else {
	node.next = queue.peek();
	}
	if (node.left != null) {
	queue.offer(node.left);
	}
	if (node.right != null) {
	queue.offer(node.right);
	}
	}
	}
	}
	class TreeLinkNode {
	TreeLinkNode left;
	TreeLinkNode right;
	TreeLinkNode next;
	}
	}

view raw ConnectRightPointers.java hosted with ❤ by GitHub