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Merge pull request #550 from KelvinG-611/98. Validate-Binary-Search-Tree
98. validate binary search tree
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@ -337,6 +337,8 @@ class Solution {
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```
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Python:
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**递归** - 利用BST中序遍历特性,把树"压缩"成数组
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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@ -344,29 +346,56 @@ Python:
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# self.val = val
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# self.left = left
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# self.right = right
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# 递归法
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class Solution:
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def isValidBST(self, root: TreeNode) -> bool:
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res = [] //把二叉搜索树按中序遍历写成list
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def buildalist(root):
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if not root: return
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buildalist(root.left) //左
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res.append(root.val) //中
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buildalist(root.right) //右
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return res
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buildalist(root)
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return res == sorted(res) and len(set(res)) == len(res) //检查list里的数有没有重复元素,以及是否按从小到大排列
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# 思路: 利用BST中序遍历的特性.
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# 中序遍历输出的二叉搜索树节点的数值是有序序列
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candidate_list = []
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def __traverse(root: TreeNode) -> None:
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nonlocal candidate_list
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if not root:
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return
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__traverse(root.left)
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candidate_list.append(root.val)
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__traverse(root.right)
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def __is_sorted(nums: list) -> bool:
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for i in range(1, len(nums)):
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if nums[i] <= nums[i - 1]: # ⚠️ 注意: Leetcode定义二叉搜索树中不能有重复元素
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return False
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return True
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__traverse(root)
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res = __is_sorted(candidate_list)
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return res
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```
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# 简单递归法
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**递归** - 标准做法
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```python
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class Solution:
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def isValidBST(self, root: TreeNode) -> bool:
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def isBST(root, min_val, max_val):
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if not root: return True
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if root.val >= max_val or root.val <= min_val:
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# 规律: BST的中序遍历节点数值是从小到大.
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cur_max = -float("INF")
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def __isValidBST(root: TreeNode) -> bool:
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nonlocal cur_max
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if not root:
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return True
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is_left_valid = __isValidBST(root.left)
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if cur_max < root.val:
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cur_max = root.val
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else:
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return False
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return isBST(root.left, min_val, root.val) and isBST(root.right, root.val, max_val)
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return isBST(root, float("-inf"), float("inf"))
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is_right_valid = __isValidBST(root.right)
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return is_left_valid and is_right_valid
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return __isValidBST(root)
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```
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```
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# 迭代-中序遍历
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class Solution:
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def isValidBST(self, root: TreeNode) -> bool:
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