[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，而且左右子树都是BST。那么以R为根的树也是BST

因此代码例如以下：

/**
 * 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;
    }
};