面试题 04.05. 合法二叉搜索树 #
Difficulty: 中等
实现一个函数，检查一棵二叉树是否为二叉搜索树。
示例 1:
输入: 2 / \ 1 3输出: true
示例 2:
输入: 5 / \ 1 4 / \ 3 6输出: false解释: 输入为: [5,1,4,null,null,3,6]。 根节点的值为 5 ，但是其右子节点值为 4 。
题解 #
题解一：BSF 遍历（Stack 数据结构） #
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
Integer currValue = null;
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
if (curr != null) {
stack.push(curr);
curr = curr.left;
} else {
curr = stack.pop();
if (currValue != null && curr.val <= currValue) {
return false;
}
currValue = curr.val;
curr = curr.right;
}
}
return true;
}
}
题解二：递归求解 #
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean isValidBST(TreeNode root, long min, long max) {
return root == null || (root.val > min && root.val < max &&
isValidBST(root.left, min, root.val) &&
isValidBST(root.right, root.val, max));
}
}