@@ -0,0 +1,85 @@
+#include<vector>
+#include<algorithm>
+#include<iostream>
+using namespace std;
+/**
+* Definition for a binary tree node.
+*/
+struct TreeNode {
+int val;
+TreeNode *left;
+TreeNode *right;
+TreeNode() : val(0), left(nullptr), right(nullptr) {}
+TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+};
+
+class Solution {
+public:
+void recoverTree(TreeNode* root) {
+TreeNode* first = nullptr; /* first: 记录第1个交换的点(逆序对) */
+TreeNode* second; /* second: 记录第2个交换的点 */
+TreeNode* last = nullptr; // last: 当前遍历指针所在位置的上一个node
+while (root != nullptr)
+{
+if (root->left == nullptr) /* 如果没有左子树, 就遍历当前点 */
+{
+if (last != nullptr && last->val > root->val) // 出现了逆序对
+{
+if (first == nullptr) // 如果是第1个逆序对, 就把逆序对的两个点都存下来
+{
+first = last;
+second = root;
+}
+else second = root; /* 如果是第2个逆序对, 只需把逆序对的第2个点都存下来 */
+}
+last = root; // 把当前点赋值给last, 为下一轮迭代做准备
+root = root->right;
+}
+else // 搭桥: 需要先找到前驱结点
+{
+auto p = root->left;
+while (p->right != nullptr && p->right != root) /* 如果当前结点(root)不为空且没有出现环路, 就继续向右移 */
+p = p->right;
+if (p->right == nullptr)
+{
+p->right = root; /* 搭桥 */
+root = root->left;
+}
+else
+{
+// 拆桥
+p->right = nullptr;
+if (last != nullptr && last->val > root->val) // 出现了逆序对, 下面的做法与前面相同
+{
+if (first == nullptr)
+{
+first = last;
+second = root;
+}
+else second = root;
+}
+last = root;
+root = root->right;
+}
+}
+}
+swap(first->val, second->val); // 把错误交换了的结点的值恢复回来
+}
+};
+
+// Test
+int main()
+{
+Solution sol;
+
+TreeNode* root = new TreeNode(3);
+root->left = new TreeNode(1);
+root->right = new TreeNode(4);
+root->right->left = new TreeNode(2);
+
+sol.recoverTree(root);
+cout << root->val << endl; /* 输出恢复后的根结点的值 */
+
+return 0;
+}

@@ -0,0 +1,69 @@
+#include<vector>
+#include<algorithm>
+#include<iostream>
+using namespace std;
+/**
+* Definition for a binary tree node.
+*/
+struct TreeNode {
+int val;
+TreeNode *left;
+TreeNode *right;
+TreeNode() : val(0), left(nullptr), right(nullptr) {}
+TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+};
+
+class Solution {
+TreeNode* lastSeen = new TreeNode(INT_MIN);
+TreeNode* first = nullptr;
+TreeNode* second = nullptr;
+public:
+void recoverTree(TreeNode* root)
+{
+inOrder(root);
+int tempVal = first->val;
+first->val = second->val;
+second->val = tempVal;
+}
+
+void inOrder(TreeNode* node)
+{
+if (node == nullptr) return;
+
+inOrder(node->left);
+if (node->val >= lastSeen->val)
+lastSeen = node;
+else /* 出现异常点 */
+{
+if (first == nullptr) /* 第1处异常点 */
+{
+first = lastSeen;
+second = node;
+lastSeen = node;
+}
+else /* 第2处异常点 */
+{
+second = node;
+return;
+}
+}
+inOrder(node->right);
+}
+};
+
+// Test
+int main()
+{
+Solution sol;
+
+TreeNode* root = new TreeNode(3);
+root->left = new TreeNode(1);
+root->right = new TreeNode(4);
+root->right->left = new TreeNode(2);
+
+sol.recoverTree(root);
+cout << root->val << endl; /* 输出恢复后的根结点的值 */
+
+return 0;
+}