# 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。