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?

=============

解法:来自leetcode 150题集

O(n)的解法,开一个指针数组,中序遍历,将节点指针一次存放在数组中,

然后寻找两处逆向的位置,先从前往后找第一个逆序的位置,然后从后往前寻找第二个逆序的位置,交换两个指针的值,

递归/非递归中序遍历一般需要用到栈,空间也是O(n)的,可以利用morris中序遍历的方式

=====

code:

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

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