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);
}
}