99. Recover Binary Search Tree

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本文介绍了一种在不改变二叉搜索树结构的情况下，恢复因两个节点错误交换导致的树状态的方法。通过中序遍历获取节点，然后对遍历结果进行排序来恢复树的正确顺序。

99. Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.

/**
 * 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) {
        //  采用中序遍历的思路，先得到所有节点中序遍历的结果，
        //  再将中序遍历的结果排序即可得到
        if (root) {
            vector<TreeNode*> vec;
            TreeNode* p = root;
            stack<TreeNode*> sta;
            while(p || !sta.empty()) {
                if (p) {
                    sta.push(p);
                    p = p -> left;
                } else {
                    vec.push_back(sta.top());
                    p = sta.top() -> right;
                    sta.pop();
                }
            }
            
            int size = vec.size();
            for (int i = 0; i < size -1; i++) {
                for (int j = i + 1; j < size; j++) {
                    if (vec[i] -> val > vec[j] -> val) {
                        int temp = vec[i] -> val;
                        vec[i] -> val = vec[j] -> val;
                        vec[j] -> val = temp;
                    }
                }
            }
        }
    }
};