[剑指Offer] Day 28: Searching and Backtracking

Authored by Tony Feng

Created on May 9th, 2022

Last Modified on May 9th, 2022

Task 1 - Q37. 序列化二叉树

Question

请实现两个函数，分别用来序列化和反序列化二叉树。你需要设计一个算法来实现二叉树的序列化与反序列化。这里不限定你的序列 / 反序列化算法执行逻辑，你只需要保证一个二叉树可以被序列化为一个字符串并且将这个字符串反序列化为原始的树结构。

Solution

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class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

from collections import deque
class Codec:
    def serialize(self, root):
        """Encodes a tree to a single string. 
        :type root: TreeNode
        :rtype: str
        """
        if not root: return "[]"
        ans = []
        queue = deque()
        queue.append(root)
        while queue: # BFS
            node = queue.popleft()
            if node:
                ans.append(str(node.val))
                queue.append(node.left)
                queue.append(node.right)
            else:
                ans.append("null")
        return '[' + ",".join(ans) + ']'

    def deserialize(self, data):
        """Decodes your encoded data to tree.
        :type data: str
        :rtype: TreeNode
        """
        if not data or data == "[]": return None
        data = data.strip('[').strip(']').split(',')
        root = TreeNode(int(data[0]))
        queue = deque()
        queue.append(root)
        i = 1

        while queue:
            node = queue.popleft()
            if data[i] != "null": # Check the left node
                node.left = TreeNode(int(data[i]))
                queue.append(node.left)
            i += 1
            if data[i] != "null": # Check the right node
                node.right = TreeNode(int(data[i]))
                queue.append(node.right)
            i += 1

        return root

Explanation

deserialize and serialize:
- Time Complexity: O(N)
- Space Complexity: O(N)

Task 2 - Q38. 字符串的排列

Question

输入一个字符串，打印出该字符串中字符的所有排列。输入：s = "abc"; 输出：["abc","acb","bac","bca","cab","cba"]

Solution

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class Solution:
    def permutation(self, s: str) -> List[str]:
        ans = set()    # Avoid duplicates
        visited = [0 for _ in range(len(s))]

        def back(path):
            if len(path) == len(s):
                ans.add(path[:])
                return

            for i in range(0, len(s)):
                if visited[i] == 0:
                    visited[i] = 1
                    back(path + s[i])
                    visited[i] = 0   # Go back

        back("")
        return list(ans)

Explanation

Time Complexity: O(N! * N)
- DFS: N * (N-1) * (N-2) * … * 1
- Set -> List: O(N)
Space Complexity: O(N²)