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?

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

中序遍历整棵树，bst的节点值则是有序的，如果发现前驱节点值大于当前节点值，那么就是异常情况

异常情况有两种，一种是前后相连的节点，另一种是不相连的节点，代码如下：

/**
 * 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 recoverTree(TreeNode *root) {
        if( !root ) return ;
        stack<TreeNode*> st;
        TreeNode* cur = root;
        while( cur ) {
            st.push(cur);
            cur = cur->left;
        }
        TreeNode* pre = 0;
        TreeNode* fnode = 0;
        TreeNode* snode = 0;
        while( !st.empty() ) {
            cur = st.top();
            st.pop();
            if( pre && pre->val > cur->val ) {  //前驱值大于当前节点值时则为异常情况
                if( !fnode ) fnode = pre, snode = cur;  //可能出现出错的两个节点刚好是前后连续的
                else {
                    snode = cur;    //再次找到前驱值大于当前节点值，那么他们就不是相连的
                    break;
                }
            }
            pre = cur;
            cur = cur->right;
            while( cur ) {
                st.push(cur);
                cur = cur->left;
            }
        }
        swap(fnode->val, snode->val);   //交换两节点的值
    }
};