[LeetCode]Recover Binary Search Tree

题目描述:(链接)

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?

解题思路:

空间复杂度是O(n), 利用中序遍历，将节点保存在一个数组中，然后遍历数组，找出错位的两个节点，交换值。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        vector<TreeNode *> cache1;
        stack<TreeNode *> cache2;
        TreeNode *p = root;
        
        while (!cache2.empty() || p != nullptr) {
            if (p != nullptr) {
                cache2.push(p);
                p = p->left;
            } else {
                p = cache2.top();
                cache2.pop();
                cache1.push_back(p);
                p = p->right;
            }
        }
        
        int len = cache1.size();
        TreeNode *first;
        for (int i = 0; i < len - 1; ++i) {
            TreeNode *tmp1 = cache1[i];
            TreeNode *tmp2 = cache1[i + 1];
            if (tmp1->val > tmp2->val) {
                first = tmp1;
                break;
            }
        }
        
        TreeNode *second;
        for (int i = len - 1; i > 0; --i) {
            TreeNode *tmp1 = cache1[i];
            TreeNode *tmp2 = cache1[i - 1];
            if (tmp1->val < tmp2->val) {
                second = tmp1;
                break;
            }
        }
        
        if (first != nullptr && second != nullptr) {
            swap(first->val, second->val);
        }
    }
};