[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?

主要思路：

使用树的递归中序遍历，当出现当前值比前一个小的时候，就存在不合法的节点。用pre存中序遍历时当前节点的前一个节点，方便值的大小对比。用p,q记录两个不合法序列的位置，p指向较小的值，q指向较大的值。

/**
 * 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 *p, *q, *pre;
    void recoverTree(TreeNode *root) {
        p = q = pre = nullptr;
        inorder(root);
        swap(p->val, q->val);
    }
    void inorder(TreeNode *root)
    {
        if(root == nullptr)  return;
        if(root->left)  inorder(root->left);
        if(pre && pre->val > root->val)
        {
            if(p == nullptr)
                p = pre;
            q = root;
        }
        pre = root;
        if(root->right) inorder(root->right);
    }
};

参考：http://www.cnblogs.com/tgkx1054/archive/2013/05/24/3096830.html