zoukankan      html  css  js  c++  java
  • leetcode-98-Validate Binary Search Tree

    leetcode-98-Validate Binary Search Tree

    98. Validate Binary Search Tree

    Description Submission Solutions Add to List

    • Total Accepted: 141105
    • Total Submissions: 626297
    • Difficulty: Medium
    • Contributors: Admin

    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:

        2
       / 
      1   3
    
    Binary tree [2,1,3], return true.

    Example 2:

        1
       / 
      2   3
    
    Binary tree [1,2,3], return false.

    Subscribe to see which companies asked this question.

    (1), 使用递归解决。 

    (2), 没想到测试用例这么极端, 居然就一定要到 int 的范围尽头, 记住 int 的范围 [-2147483648, 2147483647]

    /**
     * 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 Judge(TreeNode* root, int leftval, int rightval){
            if(root == NULL){
                return true; 
            }
            if(root->val < leftval || root->val > rightval){
                return false; 
            }
            if(root->left && root->right){
                if(root->left->val < root->val && root->right->val > root->val){
                    return Judge(root->left, leftval, root->val-1) && Judge(root->right, root->val+1, rightval); 
                }else{
                    return false; 
                }
            }else if(root->left != NULL){
                if(root->left->val < root->val){
                    return Judge(root->left, leftval, root->val-1); 
                }else{
                    return false; 
                }
            }else if(root->right != NULL){
                if(root->right->val > root->val){
                    return Judge(root->right, root->val+1, rightval); 
                }else{
                    return false; 
                }
            }else{
                return true; 
            }
        }
        
        bool isValidBST(TreeNode* root) {
            return Judge(root, -2147483648, 2147483647); 
        }
    };
    

      

  • 相关阅读:
    字节码插桩技术
    排序算法
    oracle创建简单存储过程示例
    tomcat短连接与长连接的配置
    从linux到zookeeper
    拱卒人生
    集合运算
    读取properties
    oracle的那些事
    WebService
  • 原文地址:https://www.cnblogs.com/zhang-yd/p/6399172.html
Copyright © 2011-2022 走看看