深度优先搜索DFS(前序遍历):力扣114. 二叉树展开为链表

1、题目描述:

深度优先搜索DFS(前序遍历):力扣114. 二叉树展开为链表_第1张图片

2、题解:

方法1:递归 先序遍历

先设置一个列表preorderlist,用来存储先序遍历的结果
然后先序遍历
然后,一次扫描,建立链表。
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        #先序遍历
        preorderlist = list()
        def preorderTreversal(root):
            if root:
                preorderlist.append(root)
                preorderTreversal(root.left)
                preorderTreversal(root.right)
        preorderTreversal(root)
        size = len(preorderlist)
        for i in range(1,size):
            prev,curr = preorderlist[i - 1],preorderlist[i]
            prev.left = None
            prev.right = curr

迭代

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        #迭代
        preorderlist = []
        stack = []
        node = root
        while node or stack:
            while node:
                preorderlist.append(node)
                stack.append(node)
                node = node.left
            node = stack.pop()
            node = node.right
        size = len(preorderlist)
        for i in range(1,size):
            prev,curr = preorderlist[i - 1],preorderlist[i]
            prev.left = None
            prev.right = curr

方法2:先序遍历的迭代法
如果仔细琢磨,发现方法1并不是原地
思路:

1、让root的左子树放到root的右子树处,
2、把root的原来的右子树放到root原左子树的最右结点处
3、更新root:root =  root.right
具体的:
循环:
    如果root没有左节点:
        直接让root往右节点走,也就考虑root的下一个节点
    否则:
        找左子树的最右结点pre
        让右子树挂到pre处
        让左子树插入到右子树
        令原左节点为空
        更新root	

python代码如下:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        #迭代
        while root:
            if root.left == None:
                root = root.right
            else:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
                root = root.right

3、复杂度分析:

方法1:
时间复杂度:O(N),N为结点数
空间复杂度:O(N)
方法2:
时间复杂度:O(N)
空间复杂度:O(1)

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