Validate Binary Search Tree

Validate Binary Search Tree

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.

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解题技巧：

对于一棵二叉搜索树，对其中序遍历后得到的结果是有序的。因此，我们可以先中序遍历这棵树，然后判断遍历的结果是否有序。

代码：

vector<int> res;void Inorder(TreeNode* root){    if(root == NULL) return;    Inorder(root->left);    res.push_back(root->val);    Inorder(root->right);}bool isValidBST(TreeNode* root) {       if(root == NULL) return true;    Inorder(root);    if(res.size() == 1) return true;    for(int i = 0; i < res.size() - 1; i ++)    {        if(res[i] > res[i+1]) return false;    }    return true;}