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题目

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

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

思路

通过递归,判断当前节点的左节点是否小于当前节点,右节点是否大于当前节点。这里需要注意的是子树所有节点都要满足条件。

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        return helper(root, null, null);
    }

    public boolean helper(TreeNode node, TreeNode min, TreeNode max) {
        if (node == null) return true;

        if (min != null && node.val <= min.val) return false;
        if (max != null && node.val >= max.val) return false;

        if (!helper(node.left, min, node)) return false;
        if (!helper(node.right, node, max)) return false;

        return true;
    }
}

CPP

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return helper(root, NULL, NULL);
    }
 
    bool helper(TreeNode* node, TreeNode* min, TreeNode* max) {
        if (node == NULL) return true;
     
        if (max != NULL && node->val >= max->val) return false;
        if (min != NULL && node->val <= min->val) return false;
     
        if (!helper(node->left, min, node)) return false;
        if (!helper(node->right, node, max)) return false;
     
        return true;
    }
};