leetCode 98.Validate Binary Search Tree (有效二叉搜索树) 解题思路和方法

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：判断是否二叉搜索树，按照定义判断即可，比较简单。

代码如下：

/** * 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) {        if(root == null){            return true;        }        //判断左子树的值是否全比根小,右子树是否全比根大，不是则返回false        if(!isLessThan(root, root.left) || !isGreaterThan(root, root.right)){        return false;        }        //如果左右子树均为空，为真        boolean temp = true;        //判断左子树        if(root.left != null){        if(root.val > root.left.val){        temp = isValidBST(root.left);        }else{        temp = false;        }        }        //如果左子树成立，判断右右子树        if(temp && root.right != null){        if(root.right.val > root.val){        temp = isValidBST(root.right);        }else{        temp = false;        }        }        return temp;    }    /**     * 判断左子树是否全比根节点小     * @param root     * @param left     * @return     */    private boolean isLessThan(TreeNode root, TreeNode left){    if(left == null){    return true;    }    boolean temp = false;    if(left.val < root.val){    temp = isLessThan(root, left.left) && isLessThan(root, left.right);    }    return temp;    }    /**     * 判断右子树全比根节点大     * @param root     * @param right     * @return     */    private boolean isGreaterThan(TreeNode root, TreeNode right){    if(right == null){    return true;    }    boolean temp = false;    if(right.val > root.val){    temp = isGreaterThan(root, right.left) && isGreaterThan(root, right.right);    }    return temp;    }}

另一种比较简洁的代码方式：

/** * 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) {        if(root == null){            return true;        }        //用long是因为要考虑Integer.MIn和Max的冲突        return check(root, Long.MIN_VALUE, Long.MAX_VALUE);    }    /**     * 判断根节点的是否在左右届中间     * 是则继续递归判断左右，不是则返回false     */    private boolean check(TreeNode root, long l, long r){    if(root == null)    return true;    return root.val > l && root.val < r &&     check(root.left,l,root.val) && check(root.right, root.val, r);    }}