Description

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.

My Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        if (!root){
            return true;
        }
        vector<int> temp;
        inorder(root, temp);
        for (int i = 0; i < temp.size()-1; i++){
            if(temp[i] < temp[i+1]){
                continue;
            }else{
                return false;
            }
        }
        return true;
    }
    
    void inorder(TreeNode* root, vector<int> &temp){
        if (root == NULL){
            return;
        }
        stack<TreeNode*> s;
        // TreeNode* p = root;
        while(!s.empty() || root){
            if (root){
                s.push(root);
                root = root->left;
            }else{
                root = s.top();
                temp.push_back(root->val);
                s.pop();
                root = root->right;
            }
        }
    }
};

Summary

BST的中序遍历是有序的，因此，对一棵二叉树进行中序遍历，并存入vector中，只要判断vector内的元素是有序的，则此BST就是合法的，否则就不是BST。