98. 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.

/**
 * 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) {
        return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    public boolean isValid(TreeNode root, long min, long max){
        if(root == null) return true;
        return (root.val > min && root.val < max) && isValid(root.right, root.val, max)&&
            isValid(root.left, min, root.val);
    }
}

Ending condition: root == null

False condition: Not followed the definition of BST

Recursive part: Keep traversing root's left and right child to check if they follow the rule.

The reason using long instead of int here is to avoid the situation that there's only one node whose value is Integer.MAX_VALUE.

 

https://leetcode.com/problems/validate-binary-search-tree/discuss/32112/Learn-one-iterative-inorder-traversal-apply-it-to-multiple-tree-questions-(Java-Solution)

菩萨

class Solution {
    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;
    }
}

看见bst，就要想到inorder（升序），然后配合 stack食用。