[Leetcode] Validate Binary Search Tree

Validate Binary Search Tree 题解

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题目来源：https://leetcode.com/problems/validate-binary-search-tree/description/

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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

Example 1:

    2
   / 
  1   3

Binary tree [2,1,3], return true.

Example 2:

    1
   / 
  2   3

Binary tree [1,2,3], return false.

Solution

class Solution {
private:
    void inOrder(TreeNode* node, vector<int>& arr) {
        if (node != NULL) {
            inOrder(node -> left, arr);
            arr.push_back(node -> val);
            inOrder(node -> right, arr);
        }
    }
public:
    bool isValidBST(TreeNode* root) {
        if (root == NULL) return true;
        vector<int> inOrderArr;
        inOrder(root, inOrderArr);
        int i;
        int size = inOrderArr.size();
        for (i = 0; i < size - 1; i++) {
            if (inOrderArr[i] >= inOrderArr[i + 1])
                return false;
        }
        return true;
    }
};

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解题描述

这道题题意是要检查给出的一棵二叉树是不是一棵BST。上面给出的是一开始的解法，对二叉树进行中序遍历之后记录节点值在一个数组中。遍历结束后检查数组是不是单调递增。但是这种做法不仅使用了递归，会增加运行开销，而且还需要重新遍历一次节点值数组，总的来说用了2趟。

下面给出的是迭代方式（使用栈模拟递归）中序遍历，并且在遍历的过程中增加对BST的判断，少了一趟遍历。

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        if (root == NULL) return true;
        stack<TreeNode*> nodeStack;
        TreeNode* preNode = NULL;
        while (root || !nodeStack.empty()) {
            while (root) {
                nodeStack.push(root);
                root = root -> left;
            }
            root = nodeStack.top();
            nodeStack.pop();
            if (preNode != NULL && preNode -> val >= root -> val)
                return false;
            preNode = root;
            root = root -> right;
        }
        return true;
    }
};