[LeetCode] (medium) 98. Validate Binary Search Tree

https://leetcode.com/problems/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.

Example 1:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

---

递归方法：

BST的性质决定了针对每一棵子树都需要其先祖的信息作为约束

/**
 * 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 real_function(root, (long long)INT_MIN-1, (long long)INT_MAX+1);
    }
    
    bool real_function(TreeNode *root, long long minimum, long long maximum){
        if(root == NULL) return true;
        if((long long)root->val >= maximum || (long long)root->val <= minimum) return false;
        
        if(real_function(root->left, minimum, root->val) && real_function(root->right, root->val, maximum)) 
            return true;
        else 
            return false;
    }
};

 迭代方法：

本质上就是中序遍历的各个节点严格升序

/**
 * 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) {
        stack<TreeNode*> S;
        TreeNode *cur = root;
        TreeNode *pre = NULL;
            
        while(true){
            if(cur != NULL){
                S.push(cur);
                cur = cur->left;
            }else if(!S.empty()){
                cur = S.top();  S.pop();
                if(pre != NULL && pre->val >= cur->val) return false;
                pre = cur;
                cur = cur->right;
            }else{
                break;
            }
        }
        return true;
        
    }
};

---

*附加一个别人的方法，逻辑上是使用的中序遍历，但是是递归完成的，其对于全局变量的使用和穿插特别巧妙

TreeNode prev = null;
public boolean isValidBST(TreeNode root) {
	if(root == null){
		return true;
	}
	boolean left = isValidBST(root.left);
	if(prev != null && root.val <= prev.val) return false;
	prev = root;
	boolean right = isValidBST(root.right);
	return left && right;
}