98. Validate Binary Search Tree （验证二叉搜索树，中序遍历）

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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.

 

Example 1:

    2
   / \
  1   3

Binary tree [2,1,3], return true.

 

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

 

 

 

class Solution {
public:
    TreeNode* pre; // 因为后台数据有 int最小值测试用例，所以都改成了longlong最小值。 如果测试数据中有 longlong的最小值，怎么办？不可能在初始化一个更小的值了吧。 建议避免 初始化最小值，如下pre记录前一位节点：
    bool isValidBST(TreeNode* root) {
        if(root == nullptr) return true;    
        bool left =  isValidBST(root->left);
        if (pre!=nullptr && pre->val >= root->val) return false;
        pre = root;
        bool right =  isValidBST(root->right);
        return left&&right;
    }
};

 

 

 

 

利用中序遍历，因为中序遍历的结果就是从小到大排好序的结果，如果是二叉搜索树的话。

class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root==null) return true;
        TreeNode cur = root;
        TreeNode pre = null;
        Stack<TreeNode> stack  = new Stack<TreeNode>();
        while(cur!=null || !stack.isEmpty()){
            while(cur!=null){
            stack.push(cur);
            cur = cur.left;
            }
            cur = stack.pop();
            if(pre!=null && pre.val>=cur.val) return false;
            pre =cur;
            cur=cur.right;
        }
        return true;
        
    }
  
}

 

 

 

利用递归，需要将最大值最小值传下去。

class Solution {
    public boolean isValidBST(TreeNode root) {
        return test(root,Long.MIN_VALUE,Long.MAX_VALUE); 
    }
    private boolean test(TreeNode root,long min,long max){
        if(root==null) return true;
        if(root.val>=max ||root.val<=min) return false;
        return test(root.left,min,root.val) && test(root.right,root.val,max);
    }
}