LeetCode 98. Validate Binary Search Tree

98. Validate Binary Search Tree

Difficulty:: Medium

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.

Solution

Language: Java

递归法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *       int val;
 *       TreeNode left;
 *       TreeNode right;
 *       TreeNode(int x) { val = x; }
 * }
 */
class Solution {
      public boolean isValidBST(TreeNode root) {
            if (root == null) {
                  return true;
           }
            return validHelper(root, Long.MIN_VALUE, Long.MAX_VALUE);
     }
      
      private boolean validHelper(TreeNode root, long minVal, long maxVal) {
            if (root == null) {
                  return true;
           }
            if (root.val >= maxVal || root.val <= minVal) {
                  return false;
           }
            return validHelper(root.left, minVal, root.val) && validHelper(root.right, root.val, maxVal);
     }
}

中序遍历法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        Deque<TreeNode> dq = new ArrayDeque<>();
        long prev = Long.MIN_VALUE;
        TreeNode cur = root;
        while (!dq.isEmpty() || cur != null) {
            if (cur != null) {
                dq.push(cur);
                cur = cur.left;
            } else {
                cur = dq.pop();
                if (cur.val <= prev) {
                    return false;
                }
                prev = cur.val;
                cur = cur.right;
            }
        }
        return true;
    }
}