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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

1. 用最大，最小指针在遍历二叉树时候判断是否有违反定义的节点

2和3都是用一个pre指针代表中序遍历的前一个节点，如果有pre->val >= cur->val则返回false

4. Morris 中序遍历，判断方式和2，3都一样

5. 用stack辅助做中序遍历，判断方式和2，3，4都一样

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isValidBST(TreeNode *root) {        // 1        //return isValidBST(root, NULL, NULL);                // 2        /*        TreeNode dummy(INT_MIN);        TreeNode *pre = &dummy;        return isValidBST(root, pre);        */                // 3        /*        TreeNode *pre = NULL;        return isValidBST2(root, pre);        */                // 4        TreeNode *pre = NULL;        TreeNode *cur = root;        bool ret = true;        while (cur != NULL) {            if (cur->left == NULL) {                if (pre != NULL && pre->val >= cur->val) ret = false;                pre = cur;                cur = cur->right;            } else {                TreeNode *l = cur->left;                while (l->right != NULL && l->right != cur) l = l->right;                if (l->right == NULL) {                    l->right = cur;                    cur = cur->left;                } else {                    l->right = NULL;                    if (pre != NULL && pre->val >= cur->val) ret = false;                    pre = cur;                    cur = cur->right;                }            }        }        return ret;        // 5        /*        TreeNode *pre = NULL;        TreeNode *cur = root;        stack<TreeNode*> tsk;        while (cur != NULL || !tsk.empty()) {            if (cur != NULL) {                tsk.push(cur);                cur = cur->left;                continue;            }                        cur = tsk.top();            tsk.pop();            if (pre != NULL && pre->val >= cur->val) return false;            pre = cur;            cur = cur->right;        }        */            }    private:    bool isValidBST(TreeNode *node, TreeNode *minNode, TreeNode *maxNode) {        if (node == NULL) return true;        if ((minNode != NULL && node->val <= minNode->val) || (maxNode != NULL && node->val >= maxNode->val)) return false;        return isValidBST(node->left, minNode, node) && isValidBST(node->right, node, maxNode);    }        bool isValidBST(TreeNode *node, TreeNode *&prevNode) {        if (node == NULL) return true;        if (!isValidBST(node->left, prevNode) || node->val <= prevNode->val) return false;        prevNode = node;        return isValidBST(node->right, prevNode);    }        bool isValidBST2(TreeNode *node, TreeNode *&prevNode) {        if (node == NULL) return true;        if (!isValidBST2(node->left, prevNode) || (prevNode != NULL && prevNode->val >= node->val)) return false;        prevNode = node;        return isValidBST2(node->right, prevNode);    }};