leetcode 98 Validate Binary Search Tree ----- java

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.

判断一棵树是否是二叉搜索树。

判定范围即可。主要出问题的是出现在int的最大值，最小值附近。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class Solution {
    int flag1 = 0;
    int flag2 = 0;
    public boolean isValidBST(TreeNode root) {
           return isBST(root,Integer.MAX_VALUE,Integer.MIN_VALUE);
    }
    public boolean isBST(TreeNode root,int max,int min){
        if( root == null )
            return true;
        if( root.val == Integer.MAX_VALUE && max == root.val ){
            if( flag1 == 0){
                flag1 = 1;
                return isBST(root.right,max,root.val) && isBST(root.left,root.val,min);
            }
            else
                return false;
        }
        if( root.val == Integer.MIN_VALUE && min == root.val ){
            if( flag2 == 0){
                flag2 = 1;
                return isBST(root.right,max,root.val) && isBST(root.left,root.val,min);
            }
            else
                return false;
        }
        if( root.val <=min || root.val >= max)
            return false;
        return isBST(root.right,max,root.val) && isBST(root.left,root.val,min);

    }
}

所以可以稍微优化一下。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class Solution {
    public boolean isValidBST(TreeNode root) {
        return dfs(root, (long)(Integer.MIN_VALUE)-1, (long)(Integer.MAX_VALUE)+1);
    }
    private boolean dfs(TreeNode root, long gt, long lt){
        if(root == null) return true;
        if(root.val >= lt || root.val <= gt) return false;
        return dfs(root.left, gt, root.val) && dfs(root.right, root.val, lt);
    }

}