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public class IsValidBST_98 {
class TreeNode {
int val;
TreeNode leftNode;
TreeNode rightNode;
TreeNode() {}
TreeNode(int val) {
this.val = val;
}
TreeNode(int val, TreeNode leftNode, TreeNode rightNode) {
this.val = val;
this.leftNode = leftNode;
this.rightNode = rightNode;
}
}
/**
* 二叉搜索树中序遍历按从小到大的顺序排列
* @param root
* @return
*/
long pre = Long.MIN_VALUE;
public boolean isValidBST1(TreeNode root) {
// 递归终止条件
if (root == null) {
return true;
}
// 先访问左子树
if (!isValidBST1(root.leftNode)) {
return false;
}
// 判断当前节点和上个节点大小关系
if (root.val <= pre) {
return false;
}
pre = root.val;
// 再访问右子树
return isValidBST1(root.rightNode);
}
/**
* 另一种递归的思路，通过一个辅助的递归函数来实现
* @param root
* @return
*/
public boolean isValidBST2(TreeNode root) {
return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean helper(TreeNode root, Long min, Long max) {
// 递归终止条件
if (root == null) {
return true;
}
// 当前层逻辑
if (root.val <= min || root.val >= max) {
return false;
}
// 下探到下一层
return helper(root.leftNode, min, Long.valueOf(root.val)) &&
helper(root.rightNode, Long.valueOf(root.val), max);
}
}