[LeetCode] 98. Validate Binary Search Tree

Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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.

Example1:

Input: root = [2,1,3]
Output: true

Example2:

Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

所谓的搜索二叉树，就是一颗二叉树上每个子树的左子树的值都小于它，它的右子树都数值都大于它。根据这个定义，我们可以想到两个方法。
方法一：
二叉树中序遍历(左根右), 得到一个数组，这个数组单调递增，那么根据定义，就一定是搜索二叉树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        List<Integer> rst = new ArrayList<>();
        helper(root, rst);
        return isIncrease(rst);
    }
    
    private void helper(TreeNode root, List<Integer> rst) {
        if (root == null) {
            return;
        }
        helper(root.left, rst);
        rst.add(root.val);
        helper(root.right, rst);
    }
    
    private boolean isIncrease(List<Integer> rst) {
        for (int i = 0; i < rst.size() - 1; i++) {
            if (rst.get(i) >= rst.get(i + 1)) {
                return false;
            }
        }
        return true;
    }
}

第二个思路，就是用基本的递归。但是因为需要判断每层左子树是不是都小于根，右子树是不是都大于根，所以我们需要三个信息。
1）是不是平衡 => 左边的子树是不是平衡+右边子树是不是平衡+该层是不是平衡
2）左子树的最大值 => 只要最大值小于根就所有都小于
3）右子树的最小值 => 只要最小值大于根就所有都大于

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Info {
    int leftMax;
    int rightMin;
    boolean isValid;
    Info(int leftMax, int rightMin, boolean isValid) {
        this.leftMax = leftMax;
        this.rightMin = rightMin;
        this.isValid = isValid;
    }
}
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        Info rst = helper(root);
        return rst.isValid;
    }
    
    private Info helper(TreeNode root) {
        if (root == null) {
            return null;
        }
        if (root.left == null && root.right == null) {
            return new Info(root.val, root.val, true);
        }
        
     
        Info l = helper(root.left); 
        Info r = helper(root.right);

        int leftMax = root.val;
        int rightMin = root.val;
        boolean isValid = true;
        if ( l != null) {
            isValid = l.leftMax < root.val && l.isValid && isValid;
            leftMax = Math.max(l.leftMax, leftMax); 
            rightMin = Math.min(l.rightMin, rightMin);
        }
        if (r != null) {
            isValid = r.rightMin > root.val && r.isValid && isValid;
            rightMin = Math.min(r.rightMin, rightMin);
            leftMax = Math.max(r.leftMax, leftMax);
        } 
        return new Info(leftMax, rightMin, isValid);
    }
}

标签：Binary,Search,right,TreeNode,val,int,Tree,rightMin,root
来源： https://www.cnblogs.com/codingEskimo/p/15780168.html