整體思路：
函式f對該樹排序，返回值為連結串列的頭節點
f(左子樹)
將左子樹最後一個結點和root連線
f(右子樹）
將右子樹頭結點和root連線

public class Solution {
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree==null) return pRootOfTree;
        TreeNode head = Convert(pRootOfTree.left);
        TreeNode rleft = head;
        while
 
((rleft!=null)&&(rleft.right!=null))
            rleft=rleft.right;
        if(rleft!=null)
        {
            rleft.right=pRootOfTree;
            pRootOfTree.left=rleft;
        }
        TreeNode rright = Convert(pRootOfTree.right);
        if(rright!=null)
        {
            pRootOfTree.right=rright;
            rright.left = pRootOfTree;
        }
        //此處特別注意左子樹為空的情況
 

        return head==null?pRootOfTree:head;
    }
}