【Lintcode】095.Validate Binary Search Tree

题目：

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.
• A single node tree is a BST

An example:

  2
 / \
1   4
   / \
  3   5

The above binary tree is serialized as {2,1,4,#,#,3,5} (in level order).

题解：

　　按照定义，二叉搜索树的中序遍历是升序序列

Solution 1 ()

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: True if the binary tree is BST, or false
     */
    bool isValidBST(TreeNode *root) {
        TreeNode *prev = nullptr;
        return validate(root, prev);
    }

    bool validate(TreeNode *node, TreeNode* &prev) {
        if (node == nullptr) {
            return true;
        }
        if (!validate(node->left,prev)) {
            return false;
        }
        if (prev != nullptr && prev->val >= node->val) {
            return false;
        }
        prev = node;

        return validate(node->right, prev);
    }
};

　　分治法 from here

Solution 2 ()

class ResultType {
public:
        bool isBST;
        TreeNode *maxNode, *minNode;
        ResultType(): isBST(true), maxNode(nullptr), minNode(nullptr) {}
};
class Solution {
public:

    bool isValidBST(TreeNode *root) {
        ResultType result = helper(root);
        return result.isBST;
    }

    ResultType helper(TreeNode *root) {
        ResultType result;
        if (root == nullptr) {
            return result;
        }

        ResultType left = helper(root->left);
        ResultType right = helper(root->right);

        if (!left.isBST || !right.isBST) {
            result.isBST = false;
            return result;
        }
        if (left.maxNode != nullptr && left.maxNode->val >= root->val) {
            result.isBST = false;
            return result;
        }
        if (right.minNode != nullptr && right.minNode->val <= root->val) {
            result.isBST = false;
            return result;
        }

        result.isBST = true;
        result.minNode = left.minNode == nullptr ? root : left.minNode;
        result.maxNode = right.maxNode == nullptr ? root : right.maxNode;

        return result;
    }
};