98. Validate Binary Search Tree

題目：

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.

連接：http://leetcode.com/problems/validate-binary-search-tree/

題解：

一開始使用的是BST定義，結果遇到一些邊界條件會出問題，好比 Integer.MAX_VALUE, #,Integer.MAX_VALUE一類的。因此最後仍是使用了recursive的in-order traversal。 代碼依然參考了discussion， 本身要勤練習。

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode lastNode; 
    public boolean isValidBST(TreeNode root) {
        if(root == null)
            return true;
        if(!isValidBST(root.left))
            return false;
        if(lastNode != null && lastNode.val >= root.val)
            return false;
        lastNode = root;
        if(!isValidBST(root.right))
            return false;
        return true;
    }
}

 

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二刷：

和一刷同樣的方法，先創建一個輔助節點，再用in-order traversal遍歷整個樹而且進行判斷

Java:

Time Complexity - O(n)， Space Complexity - O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode lastNode;
    
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        if (!isValidBST(root.left)) {
            return false;
        }
        if (lastNode != null && lastNode.val >= root.val) {
            return false;
        }
        lastNode = root;
        if (!isValidBST(root.right)) {
            return false;
        }
        return true;
    }
}

 

三刷:

Java:

Time Complexity - O(n)， Space Complexity - O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode prev;
    
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        if (!isValidBST(root.left)) return false;
        if (prev != null && prev.val >= root.val) return false;
        prev = root;
        if (!isValidBST(root.right)) return false;
        return true;
    }
}

 

 

Reference：

https://leetcode.com/discuss/37320/o-1-space-java-solution-morris-in-order-traversal

https://leetcode.com/discuss/45425/c-simple-recursive-solution

https://leetcode.com/discuss/39567/simple-java-recursion-solution

https://leetcode.com/discuss/22234/my-java-inorder-iteration-solution

https://leetcode.com/discuss/27913/accepted-java-solution