5.1.5 Recover Binary Sear Tree

Notes:
	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?
	Solution: 1. recursive solution. O(n) space. get inorder list first.
	2. recursive solution. O(n) space. with only auxiliary two pointers.
	3. Morris inorder traversal. O(1) space. with only auxiliary two pointers.
	*/

解题思路：先对树进行中序遍历，将遍历的结果放在容器中，然后 找到两个乱序的数。

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    // first solution
    void recoverTree_1(TreeNode *root) {
        vector<TreeNode *> inorder;
        inorderTraversal(root, inorder);
        TreeNode *first = NULL, *second = NULL;
        for (int i = 1; i < inorder.size(); ++i)
        {
            if (inorder[i-1]->val < inorder[i]->val)
                continue;
            if (!first)
                first = inorder[i-1];
            second = inorder[i];
        }
        swap(first->val, second->val);
    }

    void inorderTraversal(TreeNode *root, vector<TreeNode *> &inorder) {
        if (!root) return;
        inorderTraversal(root->left, inorder);
        inorder.push_back(root);
        inorderTraversal(root->right, inorder);
    }

｝;