LeetCode-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.

中序遍历的结果是一个递增的序列，时间复杂度为O(n)

class Solution {//树 中序遍历 递归 mytip
    private TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        return isValid(root);
    }
    private boolean isValid(TreeNode root){
        if(null==root){
            return true;
        }
        if (!isValid(root.left)||(pre!=null&&root.val<=pre.val)){//注意一定是<=
            return false;
        }
        pre = root;
        return isValid(root.right);

    }
}

时间复杂度为O(n)

class Solution {//树 递归 mytip

    public boolean isValidBST(TreeNode root) {
        long max = Long.MAX_VALUE;//Integer.MAX_VALUE不可以，要扩大范围
        long min = Long.MIN_VALUE;
        return isValid(root,min,max);
    }
    private boolean isValid(TreeNode root,long min,long max){
        if(null==root){
            return true;
        }
        if(root.val<=min||root.val>=max){
            return false;
        }
        return isValid(root.left,min,root.val)&&isValid(root.right,root.val,max);

    }
}