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.
思路：BST树的中序遍历是非递减序列
/**
* 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* mis_1;
TreeNode* mis_2;
TreeNode* pre;
void recoverTreeHelper(TreeNode* root) {
if(root == NULL) {
return;
}
if(root->left != NULL) {
recoverTreeHelper(root->left);
}
if (pre != NULL && pre->val > root->val) {
if (mis_1 != NULL) {
mis_2 = root;
}
else {
mis_1 = pre;
mis_2 = root;
}
}
pre = root;
if(root->right != NULL) {
recoverTreeHelper(root->right);
}
}
void swap(TreeNode* a, TreeNode* b) {
int tmp = a->val;
a->val = b->val;
b->val = tmp;
}
void recoverTree(TreeNode *root) {
if (root == NULL) {
return;
}
mis_1 = NULL, mis_2 = NULL, pre = NULL;
recoverTreeHelper(root);
swap(mis_1,mis_2);
}
};