算法 dfs —— 将二叉树 先序遍历 转为 链表

将二叉树拆成链表

中文English

将一棵二叉树按照前序遍历拆解成为一个 假链表。所谓的假链表是说，用二叉树的 right 指针，来表示链表中的 next 指针。

Example

样例 1：

输入：{1,2,5,3,4,#,6}
输出：{1,#,2,#,3,#,4,#,5,#,6}
解释：
     1
    / \
   2   5
  / \   \
 3   4   6
 
1
\
 2
  \
   3
    \
     4
      \
       5
        \
         6

样例 2：

输入：{1}
输出：{1}
解释：
         1
         1

Challenge

不使用额外的空间耗费。

Notice

不要忘记将左儿子标记为 null，否则你可能会得到空间溢出或是时间溢出。

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""" Definition of TreeNode: class TreeNode:     def __init__(self, val):         self.val = val         self.left, self.right = None, None """   class Solution:     """     @param root: a TreeNode, the root of the binary tree     @return: nothing     """     def flatten(self, root):         # write your code here         if not root:             return         last_node = dummy = TreeNode(None)         stack = [root]         while stack:             node = stack.pop()             last_node.right = node             if node.right:                 stack.append(node.right)             if node.left:                 stack.append(node.left)             node.left, node.right = None, None             last_node = node

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class Solution:     last_node = None           """     @param root: a TreeNode, the root of the binary tree     @return: nothing     """     def flatten(self, root):         if root is None:             return                   if self.last_node is not None:             self.last_node.left = None             self.last_node.right = root                       self.last_node = root         right = root.right         self.flatten(root.left)         self.flatten(right)

　后者是使用递归的解法。但是要注意变量修改的细节。

　需要在遍历中记住上次遍历节点，根据上次的节点和当前节点进行比较而得到result的算法模板：

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class Solution():    last_node = None    result = None         def run(self, root):         self.dfs(root)         return self.result         def dfs(self):        if last_node is None:            last_node = root        else:            do_sth(last_node, root)                     dfs(root.left)          dfs(root.right)