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.
Example 1:
2
/ \
1 3
Binary tree [2,1,3], return true.
Example 2:
1
/ \
2 3
Binary tree [1,2,3], return false.
一刷
题解：
tree的题目天生就适合用recursion来求解。注意，子问题的函数自带三个参数， root, min, max. 表示子树root内所有的值都在min-max之内。注意，在顶点时，只用Integer.MIN_VALUE和Integer.MAX_VALUE是不够的，因为这样这两个值就被排除在树之外了。所以应该用long型
/**
* 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) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean isValidBST(TreeNode root, long min, long max){
if(root == null) return true;
if(root.val<=min || root.val>=max) return false;
return isValidBST(root.left, min, root.val) &&
isValidBST(root.right, root.val, max);
}
}
二刷
思路与一刷相同
/**
* 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) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean isValidBST(TreeNode root, long min, long max){
if(root == null) return true;
if(root.val<=min || root.val>=max) return false;
return isValidBST(root.left, min, root.val) &&
isValidBST(root.right, root.val, max);
}
}
