二叉搜索树的最小绝对差

关键信息：二叉搜索树表明了树有序的！遇到在二叉搜索树上求什么最值啊，差值之类的，就把它想成在一个有序数组上求最值，求差值

# 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:# 使用迭代法if not root:return stack = []result = []while root or stack:if root:stack.append(root)root = root.leftelse:node = stack.pop(-1)result.append(node.val)root = node.rightif len(result) < 2:return res = result[1] - result[0]for i in range(2, len(result)):res = min(res, result[i] - result[i - 1])return res# 使用递归来做def getMinimumDifference(self, root: Optional[TreeNode]) -> int:result = self.getMinimumDifference1(root)min_res = result[1] - result[0]for i in range(2,len(result)):min_res = min(min_res, result[i] - result[i -1])return min_resdef getMinimumDifference1(self, root: Optional[TreeNode]) -> int:if not root:return []left = self.getMinimumDifference1(root.left)    # 左right = self.getMinimumDifference1(root.right)  # 右return left + [root.val] + right    # 左加中加右

或者更加简洁一点的方法：

501. 二叉搜索树中的众数

思路：求众数首先先到的是HashMap！也就是字典！或者利用二叉搜索树是有序的特点存在字典中

# 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.dict_1 = {}self.max = float('-inf')def findMode(self, root: Optional[TreeNode]) -> List[int]:self.findMode1(root)print(self.dict_1,self.max)result = []for i,k in self.dict_1.items():if k == self.max:result.append(i)return resultdef findMode1(self, root: Optional[TreeNode]):if not root:return if root.val in self.dict_1:self.dict_1[root.val] += 1else:self.dict_1[root.val] = 1if self.dict_1[root.val] >= self.max:self.max = self.dict_1[root.val]self.findMode1(root.left)self.findMode1(root.right)

236. 二叉树的最近公共祖先

难点：出于惯性的思维，我们都是从上往下遍历的！而这道题需要我们从下往上遍历的，需要用到后续遍历！然后节点进行回溯一步一步往上传递！

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':# 使用后续遍历的方法，可以从下往上进行处理!递归结束条件！！！if root is None or root == p or root == q:return rootleft = self.lowestCommonAncestor(root.left, p , q)  # 左right = self.lowestCommonAncestor(root.right, p , q)    # 右# 中的处理逻辑！根据左右的结果判断！这个后续遍历就很重要了if left and right:return rootelif not left and right:return rightelif left and not right:return leftelse:return None #都没有出现p或q的情况的返回给上一层的内容！

或者精简版：

class Solution:def lowestCommonAncestor(self, root, p, q):if root == q or root == p or root is None:return rootleft = self.lowestCommonAncestor(root.left, p, q)right = self.lowestCommonAncestor(root.right, p, q)if left is not None and right is not None:return rootif left is None:return rightreturn left
, p, q)if left is not None and right is not None:return rootif left is None:return rightreturn left