算法课第三周Validate Binary Search Tree

    xiaoxiao2021-03-25  63

    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. Subscribe to see which companies asked this question.

    【解析】 题意:判断一个二叉树是否为二分查找树。

    何为二分查找树?1) 左子树的值都比根节点小;2) 右子树的值都比根节点大;3) 左右子树也必须满足上面两个条件。 需要注意的是,左子树的所有节点都要比根节点小,而非只是其左孩子比其小,右子树同样。这是很容易出错的一点是,很多人往往只考虑了每个根节点比其左孩子大比其右孩子小。我们对有序二叉树进行中序遍历得到的是一个有序的排列数组,从而可以进行判断。 代码如下:

    /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { List<Integer> list = new ArrayList<Integer>(); public boolean isValidBST(TreeNode root) { if (root == null) return true; if (root.left == null && root.right == null) return true; inOrderTraversal(root); for (int i = 1; i < list.size(); i++) { if (list.get(i) <= list.get(i - 1)) return false; } return true; } public void inOrderTraversal(TreeNode root) { if (root == null) return; inOrderTraversal(root.left); list.add(root.val); inOrderTraversal(root.right); } }
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