leetcode-98-Validate Binary Search Tree

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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. 
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Example 1:

    2
   / \
  1   3

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

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [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.
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利用BST树的特性，从上往下递归，记录min_node和max_node，对左子树来说，只需记录max_node，即它的父节点；同理对于右字数，只需记录min_node。然后它们满足BST关系即可。
/**
 * 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 root != NULL ? is_bst(root, NULL, NULL) : true;
    }
    bool is_bst(TreeNode* root, TreeNode* min_node, TreeNode* max_node){
        if(root == NULL)
            return true;
        if(min_node != NULL && root->val <= min_node->val || max_node != NULL && root->val >= max_node->val)
            return false;  //if false, stop and return
        return is_bst(root->left, min_node, root) && is_bst(root->right, root, max_node);
    }
};
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利用中序遍历关系。由于BST树的中序遍历是有序的，所以我们用中序遍历来做文章。
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        TreeNode* prev = NULL;
        return is_bst(root, prev);
    }
    bool is_bst(TreeNode* root, TreeNode*& prev){
        if(root == NULL) return true;
        if(!is_bst(root->left, prev)) return false;
        if(prev != NULL && prev->val >= root->val) return false;
        prev = root;
        return is_bst(root->right, prev);
    }
};
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