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:

 /**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private int lastVal = Integer.MIN_VALUE;
public boolean isValidBST(TreeNode root) {
if(root == null)
return true; if(!isValidBST(root.left))
return false; if(root.val <= lastVal)
return false;
lastVal = root.val;
if(!isValidBST(root.right))
return false;
return true;
}
}

题目的关键点是lastVal更新的时机和与root比较的时机。

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