530. 二叉搜索树的最小绝对差

提示：

二叉搜索树（Binary Search Tree，简称 BST）是一种特殊的二叉树，它满足以下性质：

排序性：对于树中的每个节点，其左子树上所有节点的值都小于该节点的值，其右子树上所有节点的值都大于该节点的值。

不重复：树中不存在两个具有相同值的节点。

二叉性：每个节点最多有两个子节点，分别称为左子节点和右子节点。

思路：通过中序将二叉树转为数组，在寻找相邻两数差的最小值

class Solution:
    def __init__(self):
        self.vec = []

    def traversal(self, root):
        if root is None:
            return
        self.traversal(root.left)
        self.vec.append(root.val)  # 将二叉搜索树转换为有序数组
        self.traversal(root.right)

    def getMinimumDifference(self, root):
        self.vec = []
        self.traversal(root)
        if len(self.vec) < 2:
            return 0
        result = float('inf')
        for i in range(1, len(self.vec)):
            # 统计有序数组的最小差值
            result = min(result, self.vec[i] - self.vec[i - 1])
        return result

迭代法：

class Solution:
    def getMinimumDifference(self, root):
        stack = []
        cur = root
        pre = None
        result = float('inf')

        while cur is not None or len(stack) > 0:
            if cur is not None:
                stack.append(cur)  # 将访问的节点放进栈
                cur = cur.left  # 左
            else:
                cur = stack.pop()
                if pre is not None:  # 中
                    result = min(result, cur.val - pre.val)
                pre = cur
                cur = cur.right  # 右

        return result

501. 二叉搜索树中的众数

递归法（版本一）利用字典

# 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
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

迭代法：

class Solution:
    def findMode(self, root):
        st = []
        cur = root
        pre = None
        maxCount = 0  # 最大频率
        count = 0  # 统计频率
        result = []

        while cur is not None or st:
            if cur is not None:  # 指针来访问节点，访问到最底层
                st.append(cur)  # 将访问的节点放进栈
                cur = cur.left  # 左
            else:
                cur = st.pop()
                if pre is None:  # 第一个节点
                    count = 1
                elif pre.val == cur.val:  # 与前一个节点数值相同
                    count += 1
                else:  # 与前一个节点数值不同
                    count = 1

                if count == maxCount:  # 如果和最大值相同，放进result中
                    result.append(cur.val)

                if count > maxCount:  # 如果计数大于最大值频率
                    maxCount = count  # 更新最大频率
                    result = [cur.val]  # 很关键的一步，不要忘记清空result，之前result里的元素都失效了

                pre = cur
                cur = cur.right  # 右

        return result

236. 二叉树的最近公共祖先

递归法

class Solution:
    def lowestCommonAncestor(self, root, p, q):
        if root == q or root == p or root is None:
            return root

        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)

        if left is not None and right is not None:
            return root

        if left is None and right is not None:
            return right
        elif left is not None and right is None:
            return left
        else: 
            return None