LeetCode: 98. Validate Binary Search Tree
题目描述
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.
题目大意: 判断给定二叉树是否是 BST(二叉搜索树)。
解题思路
- 方法一: 由于 BST 的中序遍历是递增序列, 因此只需要中序列遍历,判断是否满足要求即可。
 - 方法二: 递归地判断相关子树是否在给定的区间内(左子树的元素都大于根节点的元素,右节点的元素都小于根节点的元素)。
 
AC 代码
- 方法一:
 
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
// 由于 BST 的中序遍历是递增序列, 因此只需要中序列遍历,判断是否满足要求即可
class Solution {
private:
    // 中序遍历
    void inOrderScan(TreeNode* root, vector<int>& inOrderSequence)
    {
        if(root == nullptr) return;
        // 中序遍历左子树
        inOrderScan(root->left, inOrderSequence);
        // 读取根节点
        inOrderSequence.push_back(root->val);
        // 中序遍历右子树
        inOrderScan(root->right, inOrderSequence);
    }
    // 判断给定序列的 [beg, end) 区间是否是递增序列
    bool isAscendSequence(const vector<int>& inOrderSequence, int beg, int end)
    {
        if(beg+1 >= end) return true;
        else if(inOrderSequence[beg] < inOrderSequence[beg+1]) return isAscendSequence(inOrderSequence, beg+1, end);
        else return false;
    }
public:
    bool isValidBST(TreeNode* root) {
        // root 的中序遍历序列
        vector<int> inOrderSequence;
        inOrderScan(root, inOrderSequence);
        return isAscendSequence(inOrderSequence, 0, inOrderSequence.size());
    }
};  - 方法二:
 
/** * 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 {
private:
    // 判断给定子树的元素是否在区间 (min, max) 中
    bool isValidBSTRef(TreeNode* root, long long min = LLONG_MIN, long long max = LLONG_MAX) {
        bool bRet = false;
        do
        {
            if(root == nullptr)
            {
                bRet = true;
                break;
            }
            if(root->val <= min || root->val >= max) break;
            // 左子树的元素应该大于根节点的元素
            if(!isValidBSTRef(root->left, min, root->val))
            {
                break;
            }
            // 右子树的元素应该小于根节点的元素
            if(!isValidBSTRef(root->right, root->val, max))
            {
                break;
            }
            bRet = true;
        }while(false);
        return bRet;
    }
public:
    bool isValidBST(TreeNode* root) {
        return isValidBSTRef(root);
    }
};
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