[LeetCode] Recover Binary Search Tree

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?

这题用O(n)的辅助空间比较好做，中序遍历后排个序O(nlogn)。但要不改变树的结构来完成就比较难了。

我只能想到一个把bst转为double link list后排序，再转为bst的方法，但转为bst就没法保证依然是原来bst的结构了。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<int> val;
    vector<TreeNode* > index;
public:
    void traverse(TreeNode *node)
    {
        if (node == NULL)
            return;
            
        traverse(node->left);
        val.push_back(node->val);
        index.push_back(node);
        traverse(node->right);
    }
    
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        traverse(root);
        sort(val.begin(), val.end());
        for(int i = 0; i < val.size(); i++)
            index[i]->val = val[i];
    }
};

 空间O(1)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void treeWalk(TreeNode* root, TreeNode*& prv, TreeNode*& first, TreeNode*& second)
    {
      if(root==NULL)
         return;
      treeWalk(root->left,prv,first,second);
      if((prv!=NULL)&&(prv->val>root->val)){
          if(first==NULL)
             first=prv;
           second=root;
      }
      prv=root;
      treeWalk(root->right,prv,first,second);
    }
  
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        TreeNode* first=NULL;
        TreeNode* second=NULL;
        TreeNode* prv=NULL;
        treeWalk(root,prv,first,second);
        int tmp=first->val;
        first->val=second->val;
        second->val=tmp;
    }
};

 空间O(1)

class Solution {
public:
    void treeWalk(TreeNode* root, TreeNode*& prv, TreeNode*& first, TreeNode*& second)
    {
      if(root==NULL)
         return;
      treeWalk(root->left,prv,first,second);
      if((prv!=NULL)&&(prv->val>root->val)){
          if(first==NULL)
             first=prv;
           second=root;
      }
      prv=root;
      treeWalk(root->right,prv,first,second);
    }
 
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        TreeNode* first=NULL;
        TreeNode* second=NULL;
        TreeNode* prv=NULL;
        treeWalk(root,prv,first,second);
        int tmp=first->val;
        first->val=second->val;
        second->val=tmp;
    }
};