PKU2255 Tree Recovery 二叉树的遍历
Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:
To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).
Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!
This is an example of one of her creations:
D
/ \
/ \
B E
/ \ \
/ \ \
A C G
/
/
F
To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).
Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!
Input
The input will contain one or more test cases.
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)
Input is terminated by end of file.
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)
Input is terminated by end of file.
Output
For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).
Sample Input
DBACEGF ABCDEFG
BCAD CBAD
Sample Output
ACBFGED
CDAB
有两种思路，一种是先构建树；再后序访问输出，
前序的第一个节点必是这棵树的根节点，
而根节点在中序的位置是它的左子树和右子树的边界，
由此我们也就知道了该根节点的左子树和右子树的前序和中序；
问题则变为构建根的左子树和右子树，并把这两棵树与根节点连起来，递归调用，最终整棵树构建起来；
另一种思路是利用递归的思想，因为相对于根节点的左子树，和右子树都是一棵独立的树；
按照后序的访问顺序：左子树，右子树，根；
最终的结果也即是 左子树的后序 + 右子树的后序 + 根节点；
所以如果我们得到了一个求后序的函数；那么先求左子树的后序，再求右子树的后序，再输出根节点，即是答案；
代码
#include
<
stdio.h
>
#include
<
string
.h
>
#include
<
stdlib.h
>
struct
node{
char
value;
struct
node
*
left,
*
right;
};
void
make_tree(
char
*
pre,
char
*
mid,
struct
node
*
root)
{
int
i,len;
int
flag
=
0
,flag1
=
0
;
struct
node
*
ptr1,
*
ptr2;
char
ppre[
28
],mmid[
28
];
for
(i
=
0
;mid[i]
!=
pre[
0
];i
++
);
//
找到根节点在中序中的位置
if
(i
!=
0
)
//
判断左子树是否存在
flag
=
1
;
if
(i
!=
(
int
)strlen(mid)
-
1
)
//
判断右子树是否存在
flag1
=
1
;
if
(flag)
//
构建该根的左子树
{
ptr1
=
(
struct
node
*
)malloc(
sizeof
(
struct
node));
ptr1
->
value
=
pre[
1
];
root
->
left
=
ptr1;
//
该根与左子树建立联系
if
(flag1
==
0
)
root
->
right
=
NULL;
strncpy(ppre,pre
+
1
,i);
//
左子树的前序
strncpy(mmid,mid,i);
//
左子树的中序
ppre[i]
=
mmid[i]
=
0
;
make_tree(ppre,mmid,ptr1);
//
已知左子树的前序和中序，构建左子树
//
free(ptr1);
}
if
(flag1)
//
构建该根的右子树
{
ptr2
=
(
struct
node
*
)malloc(
sizeof
(
struct
node));
ptr2
->
value
=
pre[i
+
1
];
root
->
right
=
ptr2;
//
该根与右子树建立联系
if
(flag
==
0
)
root
->
left
=
NULL;
len
=
strlen(pre)
-
i
-
1
;
strncpy(ppre,pre
+
i
+
1
,len);
//
得到右子树的前序
strncpy(mmid,mid
+
i
+
1
,len);
//
得到右子树的中序
ppre[len]
=
mmid[len]
=
0
;
make_tree(ppre,mmid,ptr2);
//
根据它的前序和中序，构建右子树
//
free(ptr2);
}
if
(strlen(pre)
==
1
)
root
->
left
=
root
->
right
=
NULL;
}
void
visit(
struct
node
*
p)
//
后序遍历树
{
if
(p
==
NULL)
return
;
visit(p
->
left);
visit(p
->
right);
printf(
"
%c
"
,p
->
value);
}
int
main()
{
char
pre[
28
],mid[
28
];
while
(scanf(
"
%s%s
"
,pre,mid)
!=
EOF)
{
struct
node
*
root;
root
=
(
struct
node
*
)malloc(
sizeof
(
struct
node));
root
->
value
=
pre[
0
];
make_tree(pre,mid,root);
visit(root);
putchar(
'
\n
'
);
}
return
0
;
}
第二种思路（递归方法）的代码
代码
#include
<
stdio.h
>
#include
<
string
.h
>
#define
M 30
char
stack[M];
int
top;
void
Create(
char
*
pre,
char
*
mid)
{
char
ps1[M],ps2[M],
*
qs1,
*
qs2;
int
index;
if
(strlen(pre))
{
stack[top
++
]
=
pre[
0
];
index
=
strchr(mid,pre[
0
])
-
mid;
qs1
=
mid;qs2
=
mid
+
index
+
1
;
mid[index]
=
0
;
strncpy(ps1,pre
+
1
,index);
ps1[index]
=
0
;
strcpy(ps2,pre
+
index
+
1
);
Create(ps2,qs2);
Create(ps1,qs1);
}
return
;
}
int
main()
{
char
pre[M],mid[M];
while
(scanf(
"
%s%s
"
,pre,mid)
!=-
1
)
{
top
=
0
;
Create(pre,mid);
while
(top
>
0
)
printf(
"
%c
"
,stack[
--
top]);
puts(
""
);
}
return
0
;
}