[LeetCode] 98. Validate Binary Search Tree 验证二叉搜索树

2021年09月15日 阅读数:1
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Given a binary tree, determine if it is a valid binary search tree (BST).html

Assume a BST is defined as follows:java

  • 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:node

Input:
    2
   / \
  1   3
Output: true

Example 2:python

    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.

解法1: 递归Recursiveless

解法2:迭代Iterativepost

 

Java:this

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

Java:url

public boolean isValidBST(TreeNode root) {
    if(root==null)
        return true;
 
    return helper(root, Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY);
}
 
public boolean helper(TreeNode root, double low, double high){
 
    if(root.val<=low || root.val>=high)
        return false;
 
    if(root.left!=null && !helper(root.left, low, root.val)){
        return false;
    }
 
    if(root.right!=null && !helper(root.right, root.val, high)){
        return false;
    }
 
    return true;    
}  

Java: Iterativehtm

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

Java: Iterativeblog

public class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null)
            return true;
 
        LinkedList<BNode> queue = new LinkedList<BNode>();
        queue.add(new BNode(root, Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY));
        while(!queue.isEmpty()){
            BNode b = queue.poll();
            if(b.n.val <= b.left || b.n.val >=b.right){
                return false;
            }
            if(b.n.left!=null){
                queue.offer(new BNode(b.n.left, b.left, b.n.val));
            }
            if(b.n.right!=null){
                queue.offer(new BNode(b.n.right, b.n.val, b.right));
            }
        }
        return true;
    }
}
//define a BNode class with TreeNode and it's boundaries
class BNode{
    TreeNode n;
    double left;
    double right;
    public BNode(TreeNode n, double left, double right){
        this.n = n;
        this.left = left;
        this.right = right;
    }
}  

Python: wo

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        
        return self.helper(root, float('-inf'), float('inf'))
        
    def helper(self, root, mi, mx):   
        if not root:
            return True
        
        if root.val >= mx or root.val <= mi:
            return False
            
        return self.helper(root.left, mi, root.val) and self.helper(root.right, root.val, mx)

Python:

# Morris Traversal Solution
class Solution:
    # @param root, a tree node
    # @return a list of integers
    def isValidBST(self, root):
        prev, cur = None, root
        while cur:
            if cur.left is None:
                if prev and prev.val >= cur.val:
                    return False
                prev = cur
                cur = cur.right
            else:
                node = cur.left
                while node.right and node.right != cur:
                    node = node.right

                if node.right is None:
                    node.right = cur
                    cur = cur.left
                else:
                    if prev and prev.val >= cur.val:
                        return False
                    node.right = None
                    prev = cur
                    cur = cur.right

        return True  

C++:

// Recursion without inorder traversal
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        return isValidBST(root, LONG_MIN, LONG_MAX);
    }
    bool isValidBST(TreeNode *root, long mn, long mx) {
        if (!root) return true;
        if (root->val <= mn || root->val >= mx) return false;
        return isValidBST(root->left, mn, root->val) && isValidBST(root->right, root->val, mx);
    }
};

  

  

  

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