036二叉树的镜像(keep it up)
2014-11-20 22:02
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剑指offer中题目:http://ac.jobdu.com/problem.php?pid=1521
题目描述:
输入一个二叉树,输出其镜像。
输入:
输入可能包含多个测试样例,输入以EOF结束。
对于每个测试案例,输入的第一行为一个整数n(0<=n<=1000,n代表将要输入的二叉树节点的个数(节点从1开始编号)。接下来一行有n个数字,代表第i个二叉树节点的元素的值。接下来有n行,每行有一个字母Ci。
Ci=’d’表示第i个节点有两子孩子,紧接着是左孩子编号和右孩子编号。
Ci=’l’表示第i个节点有一个左孩子,紧接着是左孩子的编号。
Ci=’r’表示第i个节点有一个右孩子,紧接着是右孩子的编号。
Ci=’z’表示第i个节点没有子孩子。
输出:
对应每个测试案例,
按照前序输出其孩子节点的元素值。
若为空输出NULL。
样例输入:
样例输出:
代码:
#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 1001
#define DEFAULTDATA 0x7fffffff
typedef struct STreeNode
{
int data;
int left;
int right;
}STreeNode;
STreeNode* createNode(int vData)
{
STreeNode *Node = (STreeNode*)malloc(sizeof(STreeNode));
Node->data = vData;
Node->left = -1;
Node->right = -1;
return Node;
}
void deleteNode(STreeNode **vNode)
{
(*vNode)->left = -1;
(*vNode)->right = -1;
free(*vNode);
*vNode = NULL;
}
void allocNode(STreeNode *vNode[], int vSize)
{
if (vSize <= 0 || vSize >= MAXSIZE) return;
int i;
for (i = 1; i <= vSize; ++i)
{
vNode[i] = createNode(DEFAULTDATA);
if (vNode[i] == NULL) return;
}
}
void deleteNode(STreeNode *vNode[], int vSize)
{
if (vSize <= 0 || vSize >= MAXSIZE) return;
int i;
for (i = 1; i <= vSize; ++i)
{
deleteNode(&(vNode[i]));
}
}
void clearNode(STreeNode *vNode[], int vSize)
{
if (vSize <= 0 || vSize >= MAXSIZE) return;
int i;
for (i = 1; i <= vSize; ++i)
{
vNode[i]->data = DEFAULTDATA;
vNode[i]->left = -1;
vNode[i]->right = -1;
}
}
void createBinaryTree(STreeNode *vTreeNode[], int vN)
{
if (vN == 0) return;
int i;
int Left;
int Right;
char Op;
for (i = 1; i <= vN; ++i)
{
while (getchar() != '\n') continue;//这个地方坑爹,直接getchar()不正确
scanf("%c", &Op);
if (Op == 'd')
{
scanf("%d %d", &Left, &Right);
vTreeNode[i]->left = Left;
vTreeNode[i]->right = Right;
}
else if (Op == 'l')
{
scanf("%d", &Left);
vTreeNode[i]->left = Left;
}
else if (Op == 'r')
{
scanf("%d", &Right);
vTreeNode[i]->right = Right;
}
}
}
void sweep(int *vLeft, int *vRight)
{
int Data;
Data =*vLeft;
*vLeft =*vRight;
*vRight = Data;
}
void switchTree(STreeNode *vTreeNode[], int vRoot)
{
if (vRoot == -1) return;
if (vTreeNode[vRoot]->left == -1 && vTreeNode[vRoot]->right == -1) return;
sweep(&(vTreeNode[vRoot]->left), &(vTreeNode[vRoot]->right));
switchTree(vTreeNode, vTreeNode[vRoot]->left);
switchTree(vTreeNode, vTreeNode[vRoot]->right);
}
void prePrintTree(STreeNode *vTreeNode[], int vRoot, int &vCount, int vN)
{
if (vRoot == -1) return;
if (vCount == vN) printf("%d\n", vTreeNode[vRoot]->data);
else printf("%d ", vTreeNode[vRoot]->data);
++vCount;
prePrintTree(vTreeNode, vTreeNode[vRoot]->left, vCount, vN);
prePrintTree(vTreeNode, vTreeNode[vRoot]->right, vCount, vN);
}
int main()
{
int i;
int N;
STreeNode *TreeNode[MAXSIZE];
allocNode(TreeNode, MAXSIZE-1);
while (scanf("%d", &N) != EOF)
{
if (N == 0)
{
printf("NULL\n");
continue;
}
for (i=1; i <= N; ++i)
{
scanf("%d", &(TreeNode[i]->data));
}
createBinaryTree(TreeNode, N);
switchTree(TreeNode, 1);
int Count = 1;
prePrintTree(TreeNode, 1, Count, N);
clearNode(TreeNode, N);
}
deleteNode(TreeNode, MAXSIZE-1);
return 0;
}
/**************************************************************
Problem: 1521
User:
Language: C++
Result: Accepted
Time:0 ms
Memory:1020 kb
****************************************************************/
题目描述:
输入一个二叉树,输出其镜像。
输入:
输入可能包含多个测试样例,输入以EOF结束。
对于每个测试案例,输入的第一行为一个整数n(0<=n<=1000,n代表将要输入的二叉树节点的个数(节点从1开始编号)。接下来一行有n个数字,代表第i个二叉树节点的元素的值。接下来有n行,每行有一个字母Ci。
Ci=’d’表示第i个节点有两子孩子,紧接着是左孩子编号和右孩子编号。
Ci=’l’表示第i个节点有一个左孩子,紧接着是左孩子的编号。
Ci=’r’表示第i个节点有一个右孩子,紧接着是右孩子的编号。
Ci=’z’表示第i个节点没有子孩子。
输出:
对应每个测试案例,
按照前序输出其孩子节点的元素值。
若为空输出NULL。
样例输入:
7 8 6 10 5 7 9 11 d 2 3 d 4 5 d 6 7 z z z z
样例输出:
8 10 11 9 6 7 5
代码:
#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 1001
#define DEFAULTDATA 0x7fffffff
typedef struct STreeNode
{
int data;
int left;
int right;
}STreeNode;
STreeNode* createNode(int vData)
{
STreeNode *Node = (STreeNode*)malloc(sizeof(STreeNode));
Node->data = vData;
Node->left = -1;
Node->right = -1;
return Node;
}
void deleteNode(STreeNode **vNode)
{
(*vNode)->left = -1;
(*vNode)->right = -1;
free(*vNode);
*vNode = NULL;
}
void allocNode(STreeNode *vNode[], int vSize)
{
if (vSize <= 0 || vSize >= MAXSIZE) return;
int i;
for (i = 1; i <= vSize; ++i)
{
vNode[i] = createNode(DEFAULTDATA);
if (vNode[i] == NULL) return;
}
}
void deleteNode(STreeNode *vNode[], int vSize)
{
if (vSize <= 0 || vSize >= MAXSIZE) return;
int i;
for (i = 1; i <= vSize; ++i)
{
deleteNode(&(vNode[i]));
}
}
void clearNode(STreeNode *vNode[], int vSize)
{
if (vSize <= 0 || vSize >= MAXSIZE) return;
int i;
for (i = 1; i <= vSize; ++i)
{
vNode[i]->data = DEFAULTDATA;
vNode[i]->left = -1;
vNode[i]->right = -1;
}
}
void createBinaryTree(STreeNode *vTreeNode[], int vN)
{
if (vN == 0) return;
int i;
int Left;
int Right;
char Op;
for (i = 1; i <= vN; ++i)
{
while (getchar() != '\n') continue;//这个地方坑爹,直接getchar()不正确
scanf("%c", &Op);
if (Op == 'd')
{
scanf("%d %d", &Left, &Right);
vTreeNode[i]->left = Left;
vTreeNode[i]->right = Right;
}
else if (Op == 'l')
{
scanf("%d", &Left);
vTreeNode[i]->left = Left;
}
else if (Op == 'r')
{
scanf("%d", &Right);
vTreeNode[i]->right = Right;
}
}
}
void sweep(int *vLeft, int *vRight)
{
int Data;
Data =*vLeft;
*vLeft =*vRight;
*vRight = Data;
}
void switchTree(STreeNode *vTreeNode[], int vRoot)
{
if (vRoot == -1) return;
if (vTreeNode[vRoot]->left == -1 && vTreeNode[vRoot]->right == -1) return;
sweep(&(vTreeNode[vRoot]->left), &(vTreeNode[vRoot]->right));
switchTree(vTreeNode, vTreeNode[vRoot]->left);
switchTree(vTreeNode, vTreeNode[vRoot]->right);
}
void prePrintTree(STreeNode *vTreeNode[], int vRoot, int &vCount, int vN)
{
if (vRoot == -1) return;
if (vCount == vN) printf("%d\n", vTreeNode[vRoot]->data);
else printf("%d ", vTreeNode[vRoot]->data);
++vCount;
prePrintTree(vTreeNode, vTreeNode[vRoot]->left, vCount, vN);
prePrintTree(vTreeNode, vTreeNode[vRoot]->right, vCount, vN);
}
int main()
{
int i;
int N;
STreeNode *TreeNode[MAXSIZE];
allocNode(TreeNode, MAXSIZE-1);
while (scanf("%d", &N) != EOF)
{
if (N == 0)
{
printf("NULL\n");
continue;
}
for (i=1; i <= N; ++i)
{
scanf("%d", &(TreeNode[i]->data));
}
createBinaryTree(TreeNode, N);
switchTree(TreeNode, 1);
int Count = 1;
prePrintTree(TreeNode, 1, Count, N);
clearNode(TreeNode, N);
}
deleteNode(TreeNode, MAXSIZE-1);
return 0;
}
/**************************************************************
Problem: 1521
User:
Language: C++
Result: Accepted
Time:0 ms
Memory:1020 kb
****************************************************************/
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