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
 *
 * 跟进“在每个节点中填充下一个右指针”的问题。
 * 如果给定的树可以是任何二叉树呢？您以前的解决方案是否仍然有效？
 * 音符：
 * 您只能使用恒定的额外空间。
 * 例如，
 * 给定以下二叉树，
 *          1
 *        /  \
 *       2    3
 *      / \    \
 *     4   5    7
 * 调用函数后，树应如下所示：
 *          1 -> NULL
 *        /  \
 *       2 -> 3 -> NULL
 *      / \    \
 *     4-> 5 -> 7 -> NULL
 */

/**
 * 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
 *
 * 跟进“在每个节点中填充下一个右指针”的问题。
 * 如果给定的树可以是任何二叉树呢？您以前的解决方案是否仍然有效？
 * 音符：
 * 您只能使用恒定的额外空间。
 * 例如，
 * 给定以下二叉树，
 *          1
 *        /  \
 *       2    3
 *      / \    \
 *     4   5    7
 * 调用函数后，树应如下所示：
 *          1 -> NULL
 *        /  \
 *       2 -> 3 -> NULL
 *      / \    \
 *     4-> 5 -> 7 -> NULL
 */

public class Main45 {
    public static void main(String[] args) {

    }

    public class TreeLinkNode {
        int val;
        TreeLinkNode left, right, next;
        TreeLinkNode(int x) { val = x; }
    }

    public void connect(TreeLinkNode root) {
        if (root == null) {
            return ;
        }
        LinkedList<TreeLinkNode> link = new LinkedList<>();
        link.offer(root);
        while (!link.isEmpty()) {
            TreeLinkNode head = link.peek();
            link.poll();
            int size = link.size();
            if (head.left != null) {
                link.offer(head.left);
            }
            if (head.right != null) {
                link.offer(head.right);
            }
            while (size > 0) {
                TreeLinkNode tree = link.peek();
                link.poll();
                head.next = tree;
                head = tree;
                if (tree.left != null) {
                    link.offer(tree.left);
                }
                if (tree.right != null) {
                    link.offer(tree.right);
                }
                size--;
            }
            head.next = null;
        }
    }

}

标签：node,head,right,pointers,tree,TreeLinkNode,link,NULL
来源： https://www.cnblogs.com/strive-19970713/p/11338373.html