【LeetCode】98. Validate Binary Search Tree 解題報告

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98. Validate Binary Search Tree / Medium

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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.

Example 1:

Input: root = [2,1,3]
Output: true

Example 2:

Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node’s value is 5 but its right child’s value is 4.

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • -2^31 <= Node.val <= 2^31 - 1

Solution 1: Recursive

思路

利用 Recursive 應該是比較直覺的做法。
但要注意的是，一顆合格的 BST，左半樹的所有節點都會小於 root 節點，所以我們必須將左半的最大值傳入；右半邊也是一樣，只是傳入的是最小值。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    bool validate(TreeNode* root, TreeNode* low, TreeNode* high) {
        if(!root) return true;
        
        if((high && root->val >= high->val) || 
            (low && root->val <= low->val)) return false;
        
        return validate(root->left, low, root) && validate(root->right, root, high);
        
    }
    
    bool isValidBST(TreeNode* root) {
         return validate(root, nullptr, nullptr);
    }
};

Solution 2: Inorder Traversal is Sorted

思路

BST 的特性之一，inorder traversal 的節點值應該會是排序狀態，所以我們將其記錄下來，檢查是否是已排序狀態。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    bool isValidBST(TreeNode* root) {
        traverse(root);
        for(int i = 1; i < nums.size(); ++i) {
            if(nums[i] <= nums[i-1]) return false;
        }
        return true;
    }
    
    void traverse(TreeNode* root) {
        if(!root) return;
        traverse(root->left);
        nums.emplace_back(root->val);
        traverse(root->right);
    }
private:
    vector<int> nums;
};