題目描述
輸入一棵二叉搜尋樹，將該二叉搜尋樹轉換成一個排序的雙向連結串列。要求不能建立任何新的結點，只能調整樹中結點指標的指向。
普通的二叉樹也可以轉換成雙向連結串列，只不過不是排序的

思路：

1. 與中序遍歷相同
2. 採用遞迴，先連結左指標，再連結右指標

程式碼1，更改doubleLinkedList，最後返回list的第一個元素：

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right =
 
 None
class Solution:
    def lastElem(self, list):
        if len(list) == 0:
            return None
        else: return list[len(list) - 1]
    def ConvertCore(self, pRoot, doubleLinkedList):
        if pRoot:
            if pRoot.left:
                self.ConvertCore(pRoot.left, doubleLinkedList)
 

            pRoot.left = self.lastElem(doubleLinkedList)
            if self.lastElem(doubleLinkedList):
                self.lastElem(doubleLinkedList).right = pRoot
            doubleLinkedList.append(pRoot)
            if pRoot.right:
                self.ConvertCore(pRoot.right, doubleLinkedList)
 

    def Convert(self, pRootOfTree):
        if pRootOfTree == None:
            return None
        doubleLinkedList = []
        self.ConvertCore(pRootOfTree, doubleLinkedList)
        return doubleLinkedList[0]

程式碼2，lastListNode指向雙向連結串列中的最後一個節點，因此每次操作最後一個節點。這裡要更改值，因此採用list的形式。

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution:
    def ConvertCore(self, pRoot, lastListNode):
        if pRoot:
            if pRoot.left:
                self.ConvertCore(pRoot.left, lastListNode)
            pRoot.left = lastListNode[0]
            if lastListNode[0]:
                lastListNode[0].right = pRoot
            lastListNode[0] = pRoot
            if pRoot.right:
                self.ConvertCore(pRoot.right, lastListNode)
    def Convert(self, pRootOfTree):
        # write code here
        if pRootOfTree == None:
            return None
        lastListNode = [None]
        self.ConvertCore(pRootOfTree, lastListNode)
        while lastListNode[0].left:
            lastListNode[0] = lastListNode[0].left
        return lastListNode[0]