LeetCode之Recover Binary Search Tree

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ /*采用Morris遍历的方法，用以确定两个被打乱的地方。 参考自：https://github.com/soulmachine/leetcode*/class Solution {public:    void recoverTree(TreeNode* root) {        //if(root == nullptr || (root->left == nullptr && root->right == nullptr))        //  return;        TreeNode *first(nullptr), *second(nullptr);        TreeNode *pre(nullptr), *cur(root);        while(cur != nullptr){            if(cur->left == nullptr){//无左子树，访问该节点                detect(&first, &second, pre, cur);                pre = cur;                cur = cur->right;            }            else{//找到左子树最右的叶子节点，并将它作为当前节点前驱                TreeNode *p = cur->left;                while(p->right != nullptr && p->right != cur) p = p->right;                                if(p->right == nullptr){//构造线索                    p->right = cur;                    cur = cur->left;                }                else{//对它进行访问                    detect(&first, &second, pre, cur);                    p->right = nullptr;                    pre = cur;                    cur = cur->right;                }            }        }        swap(first->val, second->val);    }        void detect(TreeNode **first, TreeNode **second, TreeNode *pre, TreeNode *cur){        if(pre != nullptr && pre->val > cur->val){            if(*first == nullptr) *first = pre;            *second = cur;        }    }};