LeetCode--recover-binary-search-tree

最新推荐文章于 2019-08-30 11:55:08 发布

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最新推荐文章于 2019-08-30 11:55:08 发布

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分类专栏： LeetCode 文章标签： LeetCode C

本文链接：https://blog.youkuaiyun.com/wl_ss/article/details/78998409

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本文介绍两种修复二叉搜索树中错位元素的方法：一种使用O(N)空间进行中序遍历并调整；另一种采用常数空间复杂度通过单次遍历实现。适用于面试准备及算法学习。

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题目描述

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O( n ) space is pretty straight forward. Could you devise a constant space solution?

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

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

方法一：空间复杂度为O(N)

// 使用O(n)的空间存储中序遍历的值，重排后的正确的值赋给对应的节点
class Solution {
public:
void Inorder(TreeNode *root,vector<int> &vec,vector<TreeNode *> &node)
{
    if(!root)
       return;
    Inorder(root->left,vec,node);
    vec.push_back(root->val);
    node.push_back(root);
    Inorder(root->right,vec,node);
}
void recoverTree(TreeNode *root)
{
    vector<int> vec;
    vector<TreeNode *> node;
    Inorder(root,vec,node);
    sort(vec.begin(),vec.end());
    for(int i=0;i<vec.size();i++)
        node[i]->val = vec[i];
}
};

方法二：满足空间复杂度为常数
思路：设定两个指针用来指向两处错误处和一个pre指针指向中序遍历的前一个结点。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *pre;
    TreeNode *first; //指向第一个交换点
    TreeNode *second; //指向第二个交换点
    void recoverTree(TreeNode *root)
    {
        pre = NULL;
        first = NULL;
        second = NULL;
        inorder(root);
        if (first && second) swap(first->val, second->val);
    }
    void inorder(TreeNode *root)
    {
        if (!root) return;
        inorder(root->left);  //中序遍历
        if(pre && pre->val > root->val) //当前一个点大于当前点时
         {
            if(!first)
                first =pre;
            second = root;
        }
        pre = root;
        inorder(root->right);
    }
};