原題

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.
 

Reference Answer

思路分析

看到二叉樹我們首先想到需要進行遞迴來解決問題。這道題遞迴的比較巧妙。讓我們來看下面一棵樹：

    4
   / \
  1   5
     / \
    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.

對於這棵樹而言，怎樣進行遞迴呢？root.left這棵樹的所有節點值都小於root，root.right這棵樹的所有節點值都大於root。然後依次遞迴下去就可以了。例如：如果這棵樹是二叉查詢樹，那麼左子樹的節點值一定處於（負無窮，4）這個範圍內，右子樹的節點值一定處於（4，正無窮）這個範圍內。思路到這一步，程式就不難寫了。

Reference Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        return self.helper(root, float('-inf'), float('+inf'))
    
    
    def helper(self, root, min_count, max_count):
        if not root:
            return True
        if root.val <= min_count or root.val >= max_count:
            return False
        return self.helper(root.left, min_count, root.val) and self.helper(root.right, root.val, max_count)
       

C++ 版本

/**
 * 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;
        }
        
        return helper(root, LONG_MIN, LONG_MAX);
        
    }
    bool helper(TreeNode* root, long min_count, long max_count){
        if (!root){
            return true;
        }
        if (root->val <= min_count || root->val >= max_count){
            return false;
        }
        return helper(root->left, min_count, root->val) and helper(root->right, root->val, max_count);
    }
};

Note:

• 這道題本想著中序遍歷，判斷是否順序增長，但好像一直沒有作通，還是改為與根節點進行比較判別吧。

參考文獻：

[1] https://www.cnblogs.com/zuoyuan/p/3747137.html