Validate Binary Search Tree

最新推荐文章于 2021-01-26 17:17:22 发布

原创最新推荐文章于 2021-01-26 17:17:22 发布 · 398 阅读

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本文介绍了一种算法，用于验证给定的二叉树是否为有效的二叉搜索树。通过不同方法如递归检查节点值范围、使用中序遍历及Morris遍历等技巧来判断二叉树是否符合二叉搜索树的定义。

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

1. 用最大，最小指针在遍历二叉树时候判断是否有违反定义的节点

2和3都是用一个pre指针代表中序遍历的前一个节点，如果有pre->val >= cur->val则返回false

4. Morris 中序遍历，判断方式和2，3都一样

5. 用stack辅助做中序遍历，判断方式和2，3，4都一样

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        // 1
        //return isValidBST(root, NULL, NULL);
        
        // 2
        /*
        TreeNode dummy(INT_MIN);
        TreeNode *pre = &dummy;
        return isValidBST(root, pre);
        */
        
        // 3
        /*
        TreeNode *pre = NULL;
        return isValidBST2(root, pre);
        */
        
        // 4
        TreeNode *pre = NULL;
        TreeNode *cur = root;
        bool ret = true;
        while (cur != NULL) {
            if (cur->left == NULL) {
                if (pre != NULL && pre->val >= cur->val) ret = false;
                pre = cur;
                cur = cur->right;
            } else {
                TreeNode *l = cur->left;
                while (l->right != NULL && l->right != cur) l = l->right;
                if (l->right == NULL) {
                    l->right = cur;
                    cur = cur->left;
                } else {
                    l->right = NULL;
                    if (pre != NULL && pre->val >= cur->val) ret = false;
                    pre = cur;
                    cur = cur->right;
                }
            }
        }

        return ret;
        // 5
        /*
        TreeNode *pre = NULL;
        TreeNode *cur = root;
        stack<TreeNode*> tsk;
        while (cur != NULL || !tsk.empty()) {
            if (cur != NULL) {
                tsk.push(cur);
                cur = cur->left;
                continue;
            }
            
            cur = tsk.top();
            tsk.pop();
            if (pre != NULL && pre->val >= cur->val) return false;
            pre = cur;
            cur = cur->right;
        }
        */
        
    }
    
private:
    bool isValidBST(TreeNode *node, TreeNode *minNode, TreeNode *maxNode) {
        if (node == NULL) return true;
        if ((minNode != NULL && node->val <= minNode->val) || (maxNode != NULL && node->val >= maxNode->val)) return false;
        return isValidBST(node->left, minNode, node) && isValidBST(node->right, node, maxNode);
    }
    
    bool isValidBST(TreeNode *node, TreeNode *&prevNode) {
        if (node == NULL) return true;
        if (!isValidBST(node->left, prevNode) || node->val <= prevNode->val) return false;
        prevNode = node;
        return isValidBST(node->right, prevNode);
    }
    
    bool isValidBST2(TreeNode *node, TreeNode *&prevNode) {
        if (node == NULL) return true;
        if (!isValidBST2(node->left, prevNode) || (prevNode != NULL && prevNode->val >= node->val)) return false;
        prevNode = node;
        return isValidBST2(node->right, prevNode);
    }
};