98. Validate Binary Search Tree

标签：out   treenode   def   sum   str   判断   sage   binary   top   

Problem:

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 1：

Solution I (C++):

public:
    bool isValidBST(TreeNode* root) {
        TreeNode *prev = NULL;
        return is_valid(root, prev);
    }
private:
    bool is_valid(TreeNode* root, TreeNode* &top) {
        if (root == NULL)  return true;
        if (!is_valid(root->left, top))  return false;
        if (top && top->val >= root->val)  return false;
        top = root;
        return is_valid(root->right, top);
    }

性能：

Runtime: 16 ms??Memory Usage: 20.6 MB

思路 2：

判断一个节点为根节点的左儿子还是右儿子，然后与根节点比较大小。

Solution II (C++):

public:
    bool isValidBST(TreeNode* root) {
        return isValidBST(root, NULL, NULL);
    }
private:
    bool isValidBST(TreeNode* node, TreeNode* is_lchild, TreeNode* is_rchild) {
        if (!node)  return true;
        if (is_lchild && node->val >= is_lchild->val || is_rchild && node->val <= is_rchild->val)  return false;
        return isValidBST(node->left, node, is_rchild) && isValidBST(node->right, is_lchild, node);
    }

性能：

Runtime: 20 ms??Memory Usage: 20.6 MB

98. Validate Binary Search Tree

标签：out   treenode   def   sum   str   判断   sage   binary   top   

原文地址：https://www.cnblogs.com/dysjtu1995/p/12323683.html