LeetCode第231題: 2的冪

1. 如果是2的冪, 那麼二進位制應該只有一個1, 其餘為零, 利用位運算即可判斷

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        k = 0
        while n > 0:
            k += n & 1
            n = n >> 1
            if k > 1: return False
        return True

235.二叉搜尋樹的最近公共祖先

1. 先找到每個結點的搜尋路徑, 然後找到最後一個公共結點即可

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        result1 = [
 
]
        result2 = []
        self.dfs(root, p, result1)
        self.dfs(root, q, result2)
        # for i in range(0, len(result1)):
        #     print(result1[i].val)

        # print()
        # for i in range(0, len(result2)):
        #     print(result2[i].val)
        i = 0
        length1 = len
 
(result1)
        length2 = len(result2)
        while i < length1 and i < length2:
            if result1[i] != result2[i]:
                break
            i += 1
        return result1[i-1]
    def dfs(self, root, target, result):
        result.append(root)
        if root.val == target.val:
            return
        elif root.val < target.val:
            self.dfs(root.right, target, result)
        else:
            self.dfs(root.left, target, result)

LeetCode第236題: 二叉樹的最近公共祖先

1. 思路與上一題類似, 先遍歷樹找到路徑, 然後找出兩個路徑的最後相同的結點

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        result1 = []
        result2 = []
        self.dfs(root, p, result1)
        self.dfs(root, q, result2)
        i = 0
        length1 = len(result1)
        length2 = len(result2)
        while i < length1 and i < length2:
            if result1[i] != result2[i]:
                break
            i += 1
        return result1[i-1]
    def dfs(self, root, p, result):
        if root == None:
            return False
        result.append(root)
        if root == p:
            return True
        flag = self.dfs(root.left, p, result)
        if flag:
            return True
        flag = self.dfs(root.right, p, result)
        if flag:
            return True
        result.pop()
        return False