【LeetCode & 剑指offer刷题】树题13：Validate Binary Search Tree

【LeetCode & 剑指offer刷题】树题13：Validate Binary Search Tree

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

Validate Binary Search Tree

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.

---

C++

 

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

//用中序遍历迭代法做 （递归法没看懂）

//二叉查找树性质：中序遍历后，二叉查找树为升序排列

class Solution

{

public :

    bool isValidBST ( TreeNode * root )

    {

        if ( root == NULL ) return true ;

       

        stack < TreeNode *> s ;

        TreeNode * p = root ;

        TreeNode * pre = NULL ;

       

        while (! s . empty () || p )

        {

            if ( p )

            {

                s .push(p);

 

                p = p -> left ;

            }

            else

            {

                p = s . top ();

                if(pre != NULL && p->val <= pre->val) return false; //看是否为升序

                pre = p; //保存已经访问的结点

                s . pop ();

                

                p = p -> right ;

            }

        }

       

        return true ;

    }

};

 

//参考：

        while ( p != nullptr || ! s . empty ())

        {

            if ( p != nullptr ) // 当左结点不为空时

            {

                s . push ( p ); // 入栈

 

                p = p -> left ; // 指向下一个 左结点

            }

            else              // 当左结点为空时

            {

                p = s . top ();

                 path . push_back ( p -> val ); // 访问栈顶元素（父结点）

                s . pop ();         // 出栈

               

                p = p -> right ;     // 指向 右结点

            }

        }

 

 

 

posted @ 2019-01-05 19:48 wikiwen 阅读( ...) 评论( ...) 编辑 收藏