LeetCode-99-Recover Binary Search Tree

算法描述：

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

Follow up:

• A solution using O(n) space is pretty straight forward.
• Could you devise a constant space solution?

解题思路：二叉搜索树的中序遍历是从小到大的顺序。所以，中序遍历该树，并将破坏顺序的两个节点值交换。

void recoverTree(TreeNode* root) {
        if(root == nullptr ) return;
        TreeNode* first = nullptr;
        TreeNode* second = nullptr;
        TreeNode* prev = nullptr;
        stack<TreeNode*> stk; 
        TreeNode* cur = root;
        while(cur!=nullptr || !stk.empty()){
            while(cur!=nullptr){
                stk.push(cur);
                cur=cur->left;
            }
                        
            TreeNode* temp = stk.top();
            stk.pop();
            if(prev!=nullptr && prev->val > temp->val){
                if(first==nullptr) first = prev;
                second = temp;
            }
            prev= temp;
            if(temp->right!=nullptr) cur=temp->right;
        }
        int val = first->val;
        first->val = second->val;
        second->val = val;
    }