[LeetCode]Validate Binary Search Tree

题目解析:(链接)

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.

解题思路:

先用中序遍历将bst的所有节点值保存起来，然后验证：

/**
 * 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 == nullptr) {
            return true;
        }
        
        inorder(root);
        
        for (int i = 0; i < cache.size() - 1; ++i) {
            if (cache[i] >= cache[i+1]) {
                return false;
            }
        }
        
        return true;
    }
    
private:
    vector<int> cache;
    void inorder(TreeNode *root) {
        if (!root) return;
        inorder(root->left);
        cache.push_back(root->val);
        inorder(root->right);
    }
    
};

错误解法：

该解法只是比较了根节点和他的左节点和右节点，没有递归比较和整个左子树和右子树，根节点应该大于所有左子树节点，小于所有右子树节点。

/**
 * 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 == nullptr) {
            return true;
        }
        
        if (root->left != nullptr && root->left->val >= root->val) {
            return false;
        }
        
        if (root->right != nullptr && root->right->val <= root->val) {
            return false;
        }
        
        return isValidBST(root->left) && isValidBST(root->right);
    }
    
};