333. Largest BST Subtree

最新推荐文章于 2025-04-23 12:20:04 发布

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最新推荐文章于 2025-04-23 12:20:04 发布

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原文链接：http://www.cnblogs.com/jxr041100/p/7885623.html

本文介绍了一种在二叉树中查找最大的二叉搜索树子树的方法，并提供了一个高效的解决方案，该方案的时间复杂度为O(n)。文章通过递归判断每个节点是否构成有效的二叉搜索树，并计算其节点数来确定最大子树。

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.

Note:
A subtree must include all of its descendants.
Here's an example:

The Largest BST Subtree in this case is the highlighted one.
The return value is the subtree's size, which is 3.

Follow up:
Can you figure out ways to solve it with O(n) time complexity?

从上而下的方法

/**
 * 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:
    int largestBSTSubtree(TreeNode* root) {
         if(!root) return 0;
         if(isValidBST(root,INT_MIN,INT_MAX))
             return countNodes(root);
         else
            return max(largestBSTSubtree(root->left),largestBSTSubtree(root->right)); 
    }
private:
    bool isValidBST(TreeNode *root,int minval,int maxval)
    {
        if(!root) return true;
        if(root->val<=minval || root->val>=maxval) return false;
        return isValidBST(root->left,minval,root->val) && isValidBST(root->right,root->val,maxval);
    }
    int countNodes(TreeNode*root)
    {
        if(!root) return 0;
        return 1+countNodes(root->left)+countNodes(root->right);
    }
};