最简单的想法，中序遍历二叉树：

• 递归中序遍历二叉树，得到一个中序遍历的序列
• 判断这个序列是不是有序

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

或者直接判断：

• 中序遍历，后面的节点要比前面遍历的节点值大

class Solution {
    // make sure it meets the BST not only at this level,
    // but also previous level
    // in-order access could produce a ascending order list
    // DFS
    bool helper(TreeNode * root) {
        if (root == NULL) {
            return true;
        }
        
        if (!helper(root->left)) {
            return false;
        }
        
        if (prev != NULL && prev->val >= root->val) {
            return false;
        }
        
        prev = root;
        
        return helper(root->right);
    }
public:
    TreeNode* prev = NULL;
    bool isValidBST(TreeNode* root) {
        return helper(root);
    }
};

转载于:https://www.cnblogs.com/fanhaha/p/7348446.html

原文链接：https://blog.csdn.net/weixin_30342827/article/details/96830375

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