[LeetCode]99.Recover Binary Search Tree

【题目】

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?

【解析】

O(n) 空间的解法是，开一个指针数组，中序遍历，将节点指针依次存放到数组里，然后寻找两处逆向的位置，

先从前往后找第一个逆序的位置，然后从后往前找第二个逆序的位置，交换这两个指针的值。

【代码】

/*********************************
*   日期：2014-12-10
*   作者：SJF0115
*   题号: Recover Binary Search Tree
*   来源：https://oj.leetcode.com/problems/recover-binary-search-tree/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <malloc.h>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;

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

void InOrder(TreeNode *root){
    if(root == NULL){
        return;
    }
    InOrder(root->left);
    cout<<root->val<<endl;
    InOrder(root->right);
}

class Solution {
public:
    void recoverTree(TreeNode *root) {
        stack<TreeNode*> stack;
        vector<TreeNode*> v;
        // 错乱节点
        TreeNode *first,*second;
        if(root == NULL){
            return;
        }
        TreeNode *p = root;
        // 中序遍历
        while(p != NULL || !stack.empty()){
            // 不空一直向左子树走
            if(p){
                stack.push(p);
                p = p->left;
            }
            // 转向右子树
            else{
                p = stack.top();
                stack.pop();
                v.push_back(p);
                p = p->right;
            }//if
        }//while
        // 从前向后遍历
        for(int i = 0;i < v.size();i++){
            if((v[i])->val > (v[i+1])->val){
                first = v[i];
                break;
            }
        }
        // 从后向前遍历
        for(int i = v.size()-1;i >= 0;i--){
            if((v[i])->val < (v[i-1])->val){
                second = v[i];
                break;
            }
        }
        // 交换错误节点
        /*int tmp;
        tmp = first->val;
        first->val = second->val;
        second->val = tmp;*/
        swap(first->val,second->val);
    }
};

//按先序序列创建二叉树
int CreateBTree(TreeNode* &T){
    char data;
    //按先序次序输入二叉树中结点的值（一个字符），‘#’表示空树
    cin>>data;
    if(data == '#'){
        T = NULL;
    }
    else{
        T = (TreeNode*)malloc(sizeof(TreeNode));
        //生成根结点
        T->val = data-'0';
        //构造左子树
        CreateBTree(T->left);
        //构造右子树
        CreateBTree(T->right);
    }
    return 0;
}

int main() {
    Solution solution;
    TreeNode* root(0);
    CreateBTree(root);
    solution.recoverTree(root);
    InOrder(root);
}