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98. Validate Binary Search Tree.md

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layout title date author categories description
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98. Validate Binary Search Tree
2018-11-14 11:52
Botao Xiao
Leetcode

Question:

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

  1. The left subtree of a node contains only nodes with keys less than the node's key.
  2. The right subtree of a node contains only nodes with keys greater than the node's key.
  3. 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.

Thinking:

  • Method:
    • 递归遍历，要确定是否为valid则要确定当前节点的值是否在范围之间。
      • 范围初始化值为Long.MAX_VALUE和Long.MIN_VALUE
      • 对于左子树而言，上限就是当前的结点。
      • 对于右子树而言，下限就是当前的结点。
      • 遍历到null（叶子结点），则说明为true，这也是退出递归的条件。
/**
 * 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 check(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    private static boolean check(TreeNode node, long min, long max){
        if(node == null) return true;
        if(node.val >= max || node.val <= min) return false;
        return check(node.left, min, node.val) && check(node.right, node.val, max);
    }
}

二刷

这道题写出来还是很快的，对于这种树的判断，大多是使用递归的方法。但是初始的最大值和最小值我还是用Integer类型的，所以会有Error。以后要将这样的情况考虑进来。

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

Third Time

  • Method 1: Binart Search Tree: Bottom to top

     /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 class Solution {
 	private static final int[] error = new int[]{-1, -1};
 	public boolean isValidBST(TreeNode root) {
 		if(root == null) return true;
 		int[] res = getRange(root);
 		if(res == error) return false;
 		return true;
 	}    
 	private int[] getRange(TreeNode node){  //will res[0] min, and res[1] max
 		int cur = node.val;
 		if(node.left != null && node.right != null){
 			int[] left = getRange(node.left);
 			int[] right = getRange(node.right);
 			if(left == error || right == error) return error;
 			if(cur <= left[1] || cur >= right[0]) return error;
 			else return new int[]{left[0], right[1]};
 		}else if(node.left != null || node.right != null){
 			TreeNode n = node.left != null ? node.left : node.right;
 			int[] range = getRange(n);
 			if(range == error) return error;
 			else if(node.left != null){
 				if(cur <= range[1]) return error;
 				else return new int[]{range[0], cur};
 			}else{
 				if(cur >= range[0]) return error;
 				else return new int[]{cur, range[1]};
 			}
 		}else return new int[]{cur, cur};
 	}
 }

  • Method 2: Top to bottom

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

Fourth Time

  • Method 1: recursion 
    /**
 * 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 isValid(root.left, Long.MIN_VALUE, root.val)
            && isValid(root.right, root.val, Long.MAX_VALUE);
    }
    private boolean isValid(TreeNode node, long min, long max){
        if(node == null) return true;
        else if(node.val <= min || node.val >= max) return false;
        else{
            return isValid(node.left, min, node.val) && isValid(node.right, node.val, max);
        }
    }
}

Amazon Session

  • Method 1: recursion, bottom to top 
    class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null) return true;
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    private boolean isValidBST(TreeNode node, long low, long high){
        if(node == null) return true;
        boolean left = isValidBST(node.left, low, node.val);
        boolean right = isValidBST(node.right, node.val, high);
        if(!left || !right) return false;
        else if(node.val <= low || node.val >= high) return false;
        return true;
    }
}