【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

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