【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>>
原题
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.
题目大意
验证二叉搜索树
解题思路
对二叉搜索树进行中序遍历,结果按顺序保存起来,对于二叉搜索树中序遍历其结果有一个从小到大的排列的序列,并且没有重重元素,由此可以判断一棵树是否是二叉搜索树。
代码实现
树结点类
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
- 1
- 2
- 3
- 4
- 5
- 6
- 1
- 2
- 3
- 4
- 5
- 6
算法实现类
import java.util.Stack;
public class Solution {
private Stack<Integer> stack;
public boolean isValidBST(TreeNode root) {
if (root == null) {
return true;
}
stack = new Stack<>();
inOrder(root);
int i = stack.pop();
int j;
while (!stack.isEmpty()) {
j = stack.pop();
if (i <= j) {
return false;
}
i = j;
}
return true;
}
/**
* 如果是一棵二叉查找树必必是有序的
* @param root
*/
public void inOrder(TreeNode root) {
if (root != null) {
inOrder(root.left);
stack.push(root.val);
inOrder(root.right);
}
}
}
来源:oschina
链接:https://my.oschina.net/u/2822116/blog/808755