LeetCode --- 98. Validate Binary Search Tree

98. Validate Binary Search Tree

Difficulty: Medium

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:

Input:
    2
   / \
  1   3
Output: true

Example 2:

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

Solution

思路1

二叉树中序遍历的结果是升序的，所以先中序遍历，将数字保存到vector中，判断vector中是否升序即可。
Language: C++

/**
 * 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 == NULL)   return true;
        inorder(root);
        for (int i = 1; i < v.size(); i++)
            if (v[i] <= v[i-1]) return false;
        return true;
    }
    void inorder(TreeNode* root){
        if (root == NULL)   return;
        inorder(root->left);
        v.push_back(root->val);
        inorder(root->right);
    }
private:
    vector v;
};

思路2

从每个节点本身出发，左孩子必须小于父节点，右孩子必须大于父节点，所有节点有一个不满足这个关系则返回false。

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return isValidBST(root, NULL, NULL);
    }
    bool isValidBST(TreeNode* root, TreeNode* leftNode, TreeNode* rightNode){
        if (root == NULL)   return true;
        if (leftNode && root->val <= leftNode->val || rightNode && root->val >= rightNode->val)
            return false;
        return isValidBST(root->left, leftNode, root) && isValidBST(root->right, root, rightNode);
    }
};