Validate Binary Search Tree -- leetcode

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.

算法一，遞歸+取值範圍

在leetcode上實際執行時間為32ms。

/**
 * 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) {
        return isValidBST(root, LLONG_MIN, LLONG_MAX);
    }
    
    bool isValidBST(TreeNode *root, int64_t start, int64_t stop) {
        return !root ||
               root &&
               root->val > start && root->val < stop &&
               isValidBST(root->left, start, root->val) &&
               isValidBST(root->right, root->val, stop);
    }
};
           

算法二，中序周遊+全局prev指針

在中序周遊時，檢查目前結點的值是否大于前一節點的值。

在leetcode上實際執行時間為18ms。

實際執行時間要好于前一個算法。也許是因為少一個參數入棧所緻。

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        TreeNode *prev = 0;
        return isValidBST(root, prev);
    }
    
    bool isValidBST(TreeNode *root, TreeNode *&prev) {
        if (!root) return true;
        if (!isValidBST(root->left, prev)) return false;
        if (prev && prev->val >= root->val) return false;
        prev = root;
        return isValidBST(root->right, prev);
    }
};