LeetCode Validate Binary Search Tree

/**

 * 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) {

        return dfs(root, 0, 0, 0);

    }

    

    bool dfs(TreeNode *root, int lower, int upper, int part) {

        if (root == NULL) return true;

        if ((part & 0x1) && lower >= root->val) return false;

        if ((part & 0x2) && upper <= root->val) return false;

        return dfs(root->left, lower, root->val, part | 0x2) 

                && dfs(root->right, root->val, upper, part | 0x1);

    }

};

dfs,初最左边和最右边的分支（前者只有上界，后者只有下界），其他节点都有一个上界和下界，将节点值和这个范围比较即可。

第二轮：

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.

还可以使用中序遍历检测遍历得到的序列是否是递增的

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

private:

    bool val_set;

    int pre_val;

public:

    bool isValidBST(TreeNode *root) {

        val_set = false;

        return inorder(root);

    }

    

    bool inorder(TreeNode* root) {

        if (root == NULL) {

            return true;

        }

        if (!inorder(root->left)) {

            return false;

        }

        if (!val_set) {

            pre_val = root->val;

            val_set = true;

        } else {

            if (pre_val >= root->val) {

                return false;

            }

            pre_val = root->val;

        }

        return inorder(root->right);

    }

};