# 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。
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume