Leetcode -- Recover Binary Search Tree

最新推荐文章于 2021-06-13 13:54:56 发布

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本文介绍了一种不改变二叉搜索树结构的情况下，修复两个被错误交换元素的方法。通过Morris中序遍历寻找并修正逆序节点，实现常数空间复杂度的解决方案。

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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?

由于不能采用栈，故而需要使用Morris 遍历，查找中序遍历中的两对逆序节点，并交换他们的值。

/**
 * 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:
    pair<TreeNode*,TreeNode*> mark;
    void detect(TreeNode* prev,TreeNode*cur)
    {
        if(prev&&prev->val>cur->val)
        {
            if(mark.first == NULL) mark.first = prev;
            mark.second = cur;
        }
    }
    void recoverTree(TreeNode* root) {
        TreeNode *prev=NULL,*cur=root;
        while(cur)
        {
            if(!cur->left)
            {
                detect(prev,cur);
                prev = cur;
                cur = cur->right;
            }
            else
            {
                auto node = cur->left;
                while(node->right!=NULL&&node->right!=cur)
                    node = node->right;
                if(node->right==NULL)
                {
                    node->right = cur;
                    cur = cur->left;
                }
                else
                {
                    detect(prev,cur);
                    node->right = NULL;
                    prev = cur;
                    cur = cur->right;
                }
            }
        }
        swap(mark.first->val,mark.second->val);
    }
};