1. 题目
2. 思路
(1) 递归
- 利用递归实现中序遍历，记录前一个结点的值，若当前结点的值不大于前一个结点的值，则返回false。
(2) 迭代
- 利用栈实现中序遍历。
(3) 递归优化
- 递归时传入子结点及子结点的取值范围，若子结点的值不在取值范围内，则返回false。
3. 代码
import java.util.Deque;
import java.util.LinkedList;
public class Test {
public static void main(String[] args) {
}
}
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 {
private long pre;
private boolean res;
public boolean isValidBST(TreeNode root) {
pre = Long.MIN_VALUE;
res = true;
dfs(root);
return res;
}
private void dfs(TreeNode root) {
if (!res || root == null) {
return;
}
dfs(root.left);
if (root.val <= pre) {
res = false;
return;
}
pre = root.val;
dfs(root.right);
}
}
class Solution1 {
public boolean isValidBST(TreeNode root) {
long pre = Long.MIN_VALUE;
Deque<TreeNode> stack = new LinkedList<>();
while (root != null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if (root.val <= pre) {
return false;
}
pre = root.val;
root = root.right;
}
return true;
}
}
class Solution2 {
public boolean isValidBST(TreeNode root) {
return recur(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean recur(TreeNode root, long min, long max) {
if (root == null) {
return true;
}
if (root.val <= min || root.val >= max) {
return false;
}
return recur(root.left, min, root.val) && recur(root.right, root.val, max);
}
}