[LeetCode] Validate Binary Search Tree

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.

解题思路：

（1）空二叉树是BST

（2）若根节点R值大于左子树的最大值，且小于右子树最小值，并且左右子树都是BST，那么以R为根的树也是BST

因此代码如下：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isValidBST(TreeNode* root) {        if(root == NULL){            return true;        }        TreeNode* leftMaxNode = getMaxNode(root->left);        TreeNode* rightMinNode = getMinNode(root->right);        if(leftMaxNode!=NULL && leftMaxNode->val >= root->val){            return false;        }        if(rightMinNode!=NULL && rightMinNode->val <= root->val){            return false;        }        return isValidBST(root->left) && isValidBST(root->right);    }        TreeNode* getMaxNode(TreeNode* root){        if(root == NULL){            return NULL;        }        TreeNode* maxNode = root;        TreeNode* leftMaxNode = getMaxNode(root->left);        TreeNode* rightMaxNode = getMaxNode(root->right);        if(leftMaxNode!=NULL && leftMaxNode->val > maxNode->val){            maxNode = leftMaxNode;        }        if(rightMaxNode!=NULL && rightMaxNode->val > maxNode->val){            maxNode = rightMaxNode;        }        return maxNode;    }        TreeNode* getMinNode(TreeNode* root){        if(root == NULL){            return NULL;        }        TreeNode* minNode = root;        TreeNode* leftMinNode = getMinNode(root->left);        TreeNode* rightMinNode = getMinNode(root->right);        if(leftMinNode!=NULL && leftMinNode->val < minNode->val){            minNode = leftMinNode;        }        if(rightMinNode!=NULL && rightMinNode->val < minNode->val){            minNode = rightMinNode;        }        return minNode;    }};