Validate Binary Search Tree
最后更新于：2022-04-02 01:11:48
# Validate Binary Search Tree
### Source
- lintcode: [(95) Validate Binary Search Tree](http://www.lintcode.com/en/problem/validate-binary-search-tree/)
~~~
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Example
An example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
~~~
### 题解1 - recursion
按照题中对二叉搜索树所给的定义递归判断，我们从递归的两个步骤出发分析：
1. 基本条件/终止条件 - 返回值需斟酌。
1. 递归步/条件递归 - 能使原始问题收敛。
终止条件好确定——当前节点为空，或者不符合二叉搜索树的定义，返回值分别是什么呢？先别急，分析下递归步试试先。递归步的核心步骤为比较当前节点的`key`和左右子节点的`key`大小，和定义不符则返回`false`, 否则递归处理。从这里可以看出在节点为空时应返回`true`, 由上层的其他条件判断。但需要注意的是这里不仅要考虑根节点与当前的左右子节点，**还需要考虑左子树中父节点的最小值和右子树中父节点的最大值。**否则程序在`[10,5,15,#,#,6,20]` 这种 case 误判。
由于不仅需要考虑当前父节点，还需要考虑父节点的父节点... 故递归时需要引入上界和下界值。画图分析可知对于左子树我们需要比较父节点中最小值，对于右子树则是父节点中的最大值。又由于满足二叉搜索树的定义时，左子结点的值一定小于根节点，右子节点的值一定大于根节点，故无需比较所有父节点的值，使用递推即可得上界与下界，这里的实现非常巧妙。
### C++ - long long
~~~
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: True if the binary tree is BST, or false
*/
bool isValidBST(TreeNode *root) {
if (root == NULL) return true;
return helper(root, LLONG_MIN, LLONG_MAX);
}
bool helper(TreeNode *root, long long lower, long long upper) {
if (root == NULL) return true;
if (root->val <= lower || root->val >= upper) return false;
bool isLeftValidBST = helper(root->left, lower, root->val);
bool isRightValidBST = helper(root->right, root->val, upper);
return isLeftValidBST && isRightValidBST;
}
};
~~~
### C++ - without long long
~~~
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: True if the binary tree is BST, or false
*/
bool isValidBST(TreeNode *root) {
if (root == NULL) return true;
return helper(root, INT_MIN, INT_MAX);
}
bool helper(TreeNode *root, int lower, int upper) {
if (root == NULL) return true;
if (root->val <= lower || root->val >= upper) {
bool right_max = root->val == INT_MAX && root->right == NULL;
bool left_min = root->val == INT_MIN && root->left == NULL;
if (!(right_max || left_min)) {
return false;
}
}
bool isLeftValidBST = helper(root->left, lower, root->val);
bool isRightValidBST = helper(root->right, root->val, upper);
return isLeftValidBST && isRightValidBST;
}
};
~~~
### Java
~~~
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: True if the binary tree is BST, or false
*/
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean helper(TreeNode root, long lower, long upper) {
if (root == null) return true;
// System.out.println("root.val = " + root.val + ", lower = " + lower + ", upper = " + upper);
// left node value < root node value < right node value
if (root.val >= upper || root.val <= lower) return false;
boolean isLeftValidBST = helper(root.left, lower, root.val);
boolean isRightValidBST = helper(root.right, root.val, upper);
return isLeftValidBST && isRightValidBST;
}
}
~~~
### 源码分析
为避免节点中出现整型的最大最小值，引入 long 型进行比较。有些 BST 的定义允许左子结点的值与根节点相同，此时需要更改比较条件为`root.val > upper`. C++ 中 long 可能与 int 范围相同，故使用 long long. 如果不使用比 int 型更大的类型，那么就需要在相等时多加一些判断。
### 复杂度分析
递归遍历所有节点，时间复杂度为 O(n)O(n)O(n), 使用了部分额外空间，空间复杂度为 O(1)O(1)O(1).
### 题解2 - iteration
联想到二叉树的中序遍历。TBD
### Reference
- [LeetCode: Validate Binary Search Tree 解题报告 - Yu's Garden - 博客园](http://www.cnblogs.com/yuzhangcmu/p/4177047.html) - 提供了4种不同的方法，思路可以参考。
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