力扣hot100 - 114、二叉树展开为链表

题目：思路一：通过先序遍历将元素放到一个数组中，再去遍历数组同时构造一棵二叉树。

代码略

思路二：先递归地把左子树展开为一个链表，再记录一下右子树，左子树移到右子树位置，把记录的右子树放到左子树右节点。

/**
 * 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) {
        if(root == null){
            return;
        }
        flatten(root.left);
        flatten(root.right);
        TreeNode trmp = root.right;
        root.right = root.left;
        root.left = null;
        TreeNode cur = root;
        while(cur != null){
             if( cur.right == null){
                cur.right = trmp;
                break;
            }
            cur = cur.right;
        
        }
        return;
    }
}