530. 二叉搜索树的最小绝对差

题目

思路与解法

第一想法： 一个二叉搜索树的最小绝对差，从根结点看，它的结点与它的最小差值一定出现在 左子树的最右结点（左子树最大值）和右子树的最左结点（右子树的最小值）。
这样搞复杂了，直接层序遍历，然后后值减前值，记录最小值

# 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 getMinimumDifference(self, root: Optional[TreeNode]) -> int:
        self.inorder_list = []
        self.inorder(root)
        res = float('inf')
        for i in range(len(self.inorder_list)-1):
            cur = self.inorder_list[i+1] - self.inorder_list[i]
            res = cur if cur < res else res
        return res

    def inorder(self,root):
        if not root:
            return
        self.inorder(root.left)
        self.inorder_list.append(root.val)
        self.inorder(root.right)

501.二叉搜索树中的众数

题目

思路与解法

第一想法： 遍历

# 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 findMode(self, root: Optional[TreeNode]) -> List[int]:
        from collections import defaultdict

class Solution:
    def searchBST(self, cur, freq_map):
        if cur is None:
            return
        freq_map[cur.val] += 1  # 统计元素频率
        self.searchBST(cur.left, freq_map)
        self.searchBST(cur.right, freq_map)

    def findMode(self, root):
        freq_map = defaultdict(int)  # key:元素，value:出现频率
        result = []
        if root is None:
            return result
        self.searchBST(root, freq_map)
        max_freq = max(freq_map.values())
        for key, freq in freq_map.items():
            if freq == max_freq:
                result.append(key)
        return result

236. 二叉树的最近公共祖先

题目

思路与解法

carl的讲解： 如果左右都有返回值，那就是它。不可能出现第二个左右都有返回值的节点，因为本题树中节点不重复。

# 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':
        if root == p or root == q or root == None:
            return root
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        
        if left and right:
            return root
        if not left and right:
            return right
        elif left and not right:
            return left
        else:
            return None