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思路源自 

leetcode-树篇 114题 二叉树展开为链表

 对于当前结点来说，先把左子树全部拉平成链表，再把右子树也全部拉平成链表

然后记录当前结点的右子树起始点temp，把当前结点右指针指向左子树，左子树的最右下边的结点的右指针指向一开始记录的temp，最后就拉平成链表了

/*** 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) {flatten(root.left);flatten(root.right);TreeNode temp = root.right;root.right=root.left;root.left = null;while (root.right!=null) root=root.right;root.right = temp;}}
}