给定一颗二叉搜索树，请找出其中的第k小的结点。例如， 5 / \ 3 7 /\ /\ 2 4 6 8 中，按结点数值大小顺序第三个结点的值为4。...

// ConsoleApplication2.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include "stdafx.h"
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;





struct TreeNode {
    char val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
    val(x), left(NULL), right(NULL) {
    }
};
class Solution {
public:
    TreeNode* KthNode(TreeNode* pRoot, unsigned int k)
    {
        getData(pRoot);
        if (k <= 0) return NULL;
        if (k > qu.size()) return NULL;
        for (int i = 1; i < k; i++)
        {
            qu.pop();
        }
        return qu.front();
    }

    queue qu;
    void getData(TreeNode* T)
    {
        if (T == NULL) return;
        else
        {   
            getData(T->left);
            qu.push(T);
            getData(T->right);
        }
    }

    //序列化二叉树，前序创建二叉树
    int p = -1;
    TreeNode* Deserialize(char *str) {
        if (*str == NULL) return NULL; //str为空
        TreeNode *T = NULL;
        ++p;
        if (p >= strlen(str)) return NULL; //超出str的范围
        if (str[p] == '#') return NULL; //str的值等于#
        T = new TreeNode(str[p]);
        T->left = Deserialize(str);
        T->right = Deserialize(str);
        return T;
    }
    void preOrder(TreeNode *T)
    {
        if (T == NULL) return;
        else
        {
            cout << T->val << "  ";
            preOrder(T->left);
            preOrder(T->right);
        }
    }

};

int main()
{

    //int num = 5;
    //char *ch = new char[num];
    //cout << "strlen(ch):" << strlen(ch)<val << endl;


    //so.getData(T);
    //cout << "qu 队列中的值：" << endl;
    //so.print();
    //cout << endl;


    return 0;
}

转载于:https://www.cnblogs.com/wdan2016/p/6000152.html