0114.二叉树展开为链表.java

/*
 * @lc app=leetcode.cn id=114 lang=java
 *
 * [114] 二叉树展开为链表
 *
 * https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/description/
 *
 * algorithms
 * Medium (72.95%)
 * Likes:    1293
 * Dislikes: 0
 * Total Accepted:    300.9K
 * Total Submissions: 412.4K
 * Testcase Example:  '[1,2,5,3,4,null,6]'
 *
 * 给你二叉树的根结点 root ，请你将它展开为一个单链表：
 * 
 * 
 * 展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
 * 展开后的单链表应该与二叉树 先序遍历 顺序相同。
 * 
 * 
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：root = [1,2,5,3,4,null,6]
 * 输出：[1,null,2,null,3,null,4,null,5,null,6]
 * 
 * 
 * 示例 2：
 * 
 * 
 * 输入：root = []
 * 输出：[]
 * 
 * 
 * 示例 3：
 * 
 * 
 * 输入：root = [0]
 * 输出：[0]
 * 
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * 树中结点数在范围 [0, 2000] 内
 * -100 
 * 
 * 
 * 
 * 
 * 进阶：你可以使用原地算法（O(1) 额外空间）展开这棵树吗？
 * 
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        TreeNode current = root;
        while (current != null) {
            if (current.left != null) {
                TreeNode left = current.left;
                TreeNode predecessor = left;
                while (predecessor.right != null)
                    predecessor = predecessor.right;

                predecessor.right = current.right;
                current.left = null;
                current.right = left;
            }

            current = current.right;
        }
    }
}
// @lc code=end