代码随想录算法训练营day21 | 530.二叉搜索树的最小绝对差 | 501.二叉搜索树中的众数 | 236. 二叉树的最近公共祖先

530.二叉搜索树的最小绝对差

迭代

class Solution:
    def getMinimumDifference(self, root: TreeNode) -> int:
        stack = []
        cur = root
        pre = None
        result = float('inf')
        while cur or stack:
            if cur: # 指针来访问节点，访问到最底层
                stack.append(cur)
                cur = cur.left
            else: # 逐一处理节点
                cur = stack.pop()
                if pre: # 当前节点和前节点的值的差值
                    result = min(result, abs(cur.val - pre.val))
                pre = cur
                cur = cur.right
        return result

递归

class Solution:
    def getMinimumDifference(self, root: TreeNode) -> int:
        res = []   
        r = float("inf")
        def buildaList(root):  //把二叉搜索树转换成有序数组
            if not root: return None
            if root.left: buildaList(root.left)  //左
            res.append(root.val)  //中
            if root.right: buildaList(root.right)  //右
            return res
            
        buildaList(root)
        for i in range(len(res)-1):  // 统计有序数组的最小差值
            r = min(abs(res[i]-res[i+1]),r)
        return r

 501.二叉搜索树中的众数

迭代

class Solution:
    def findMode(self, root: TreeNode) -> List[int]:
        stack = []
        cur = root
        pre = None
        maxCount, count = 0, 0
        res = []
        while cur or stack:
            if cur:  # 指针来访问节点，访问到最底层
                stack.append(cur)
                cur = cur.left
            else:  # 逐一处理节点
                cur = stack.pop()
                if pre == 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:
                    count = 1
                if count == maxCount:
                    res.append(cur.val)
                if count > maxCount:
                    maxCount = count
                    res.clear()
                    res.append(cur.val)

                pre = cur
                cur = cur.right
        return res	

递归

# 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 __init__(self):
        self.pre = TreeNode()
        self.count = 0
        self.max_count = 0
        self.result = []

    def findMode(self, root: TreeNode) -> List[int]:
        if not root: return None
        self.search_BST(root)
        return self.result
        
    def search_BST(self, cur: TreeNode) -> None:
        if not cur: return None
        self.search_BST(cur.left)
        # 第一个节点
        if not self.pre:
            self.count = 1
        # 与前一个节点数值相同
        elif self.pre.val == cur.val:
            self.count += 1 
        # 与前一个节点数值不相同
        else:
            self.count = 1
        self.pre = cur

        if self.count == self.max_count:
            self.result.append(cur.val)
        
        if self.count > self.max_count:
            self.max_count = self.count
            self.result = [cur.val]	# 清空self.result，确保result之前的的元素都失效
        
        self.search_BST(cur.right)

 主要要注意clr那步

236. 二叉树的最近公共祖先 

这个太难了听不懂呐……