leetcode[99] Recover Binary Search Tree

/**
 * 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 inorder(TreeNode *root, vector<int> &perm) // 中序遍历得到数组
    {
        if (root == NULL) return ;
        stack<TreeNode *> sta;
        
        sta.push(root);
        TreeNode *p = root -> left, *cen;
        
        while(!sta.empty())
        {
            if (p)
            {
                sta.push(p);
                p = p -> left;
            }
            else
            {
                cen = sta.top();
                sta.pop();
                perm.push_back(cen -> val);
                if (cen -> right)
                {
                    sta.push(cen -> right);
                    p = cen -> right -> left;
                }
            }
        }
    }
        void inorderC(TreeNode *root,int val1, int val2) // 中序遍历将对应值修改
    {
        if (root == NULL) return ;
        stack<TreeNode *> sta;
        
        sta.push(root);
        TreeNode *p = root -> left, *cen;
        
        while(!sta.empty())
        {
            if (p)
            {
                sta.push(p);
                p = p -> left;
            }
            else
            {
                cen = sta.top();
                sta.pop();
                if(cen -> val == val1)
                    cen -> val = val2;
                else if(cen -> val == val2)
                    cen -> val = val1;
                if (cen -> right)
                {
                    sta.push(cen -> right);
                    p = cen -> right -> left;
                }
            }
        }
    }
    void recoverTree(TreeNode *root) 
    {
        if (!root) return ;
        vector<int> perm;
        inorder(root, perm);
        int val1, val2;
        bool flag = true;
        for (int i = 0; i < perm.size(); ++i)
        {
            if (flag && i + 1 < perm.size() && perm[i] > perm[i+1]) // 第一次比后一个大的数
            {
                val1 = perm[i];
                flag = false;
            }
            if (i - 1 >= 0 && perm[i] < perm[i-1]) // 最后一个比前面小的数
                val2 = perm[i];
        }
        inorderC(root, val1, val2);
    }
};