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.
题解:BST valid 的充分必要条件是它的中序遍历是一个有序序列。
递归实现树的中序遍历,用私有变量lastVal记录上一个遍历的节点的值。在一次递归,首先递归判断左子树是否是BST,并且更新lastVal,然后将root的值跟lastVal比较,看root的值是否大于lastVal;然后递归判断右子树是否是BST。
代码如下:
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 private int lastVal = Integer.MIN_VALUE; 12 public boolean isValidBST(TreeNode root) { 13 if(root == null) 14 return true; 15 16 if(!isValidBST(root.left)) 17 return false; 18 19 if(root.val <= lastVal) 20 return false; 21 lastVal = root.val; 22 if(!isValidBST(root.right)) 23 return false; 24 return true; 25 } 26 }
题目的关键点是lastVal更新的时机和与root比较的时机。