HDU1710 Binary Tree Traversals
同PKU2255 Tree Recovery可以说是一道题目了
本题题目描述：链接
仿照PKU2255写的代码，一不小心给过了。
先由前序 中序建树，再后序遍历 62MS 1360K（内存有点大啊）
代码
#include
<
stdio.h
>
#include
<
stdlib.h
>
struct
node{
int
value;
struct
node
*
left,
*
right;
};
int
index,n;
void
make_tree(
int
*
pre,
int
plen,
int
*
in
,
int
ilen,
struct
node
*
root)
{
int
flag
=
0
,flag1
=
0
;
int
i;
struct
node
*
ptr1,
*
ptr2;
for
(i
=
0
;
in
[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
!=
plen
-
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;
make_tree(pre
+
1
,i,
in
,i,ptr1);
//
已知左子树的前序和中序，构建左子树
}
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;
make_tree(pre
+
i
+
1
,plen
-
i
-
1
,
in
+
i
+
1
,ilen
-
i
-
1
,ptr2);
//
根据它的前序和中序，构建右子树
}
if
(plen
==
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);
if
(index
!=
n)
printf(
"
%d
"
,p
->
value);
else
printf(
"
%d\n
"
,p
->
value);
index
++
;
}
#define
N 1002
int
main()
{
int
pre[N],
in
[N],i;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
for
(i
=
0
;i
<
n;i
++
)
scanf(
"
%d
"
,
&
pre[i]);
for
(i
=
0
;i
<
n;i
++
)
scanf(
"
%d
"
,
&
in
[i]);
struct
node
*
root;
root
=
(
struct
node
*
)malloc(
sizeof
(
struct
node));
root
->
value
=
pre[
0
];
make_tree(pre,n,
in
,n,root);
index
=
1
;
visit(root);
}
return
0
;
}
见了队友William用深搜建树，虽然知道和上述方法是同一个道理，运用递归思想，既然写了也贴一下，做对比78MS 1364K
要理解啊，请教了William我算理解了其真谛了吧 ^_^
代码
#include
<
stdio.h
>
#include
<
stdlib.h
>
typedef
struct
node{
int
data;
int
index;
struct
node
*
left,
*
right;
}bitree,
*
Tree;
int
n,pre[
1005
],mid[
1005
];
void
dfs(Tree
&
root,
int
index)
{
if
(root
==
NULL)
{
root
=
(Tree)malloc(
sizeof
(bitree));
root
->
data
=
mid[index];
root
->
index
=
index;
root
->
left
=
NULL;
root
->
right
=
NULL;
}
else
{
if
(index
<
root
->
index)
dfs(root
->
left,index);
else
dfs(root
->
right,index);
}
}
void
creatbitree(Tree
&
root)
{
int
i,j,index;
root
=
(Tree)malloc(
sizeof
(bitree));
for
(i
=
1
;i
<=
n;i
++
)
{
if
(mid[i]
==
pre[
1
])
{
root
->
data
=
pre[
1
];
root
->
index
=
i;
root
->
left
=
NULL;
root
->
right
=
NULL;
break
;
}
}
index
=
i;
for
(i
=
2
;i
<=
n;i
++
)
{
for
(j
=
1
;j
<=
n;j
++
)
if
(mid[j]
==
pre[i])
{
if
(j
<
index)
dfs(root
->
left,j);
else
dfs(root
->
right,j);
break
;
}
}
}
void
post(Tree
&
root,
int
x)
{
if
(root
==
NULL)
return
;
post(root
->
left,x
+
1
);
post(root
->
right,x
+
1
);
if
(x
==
0
)
printf(
"
%d
"
,root
->
data);
else
printf(
"
%d
"
,root
->
data);
}
int
main()
{
int
i;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
Tree root;
for
(i
=
1
;i
<=
n;i
++
)
scanf(
"
%d
"
,
&
pre[i]);
for
(i
=
1
;i
<=
n;i
++
)
scanf(
"
%d
"
,
&
mid[i]);
creatbitree(root);
post(root,
0
);
printf(
"
\n
"
);
}
return
0
;
}
下面贴一个很牛的代码（至少和前两个比），46MS 160K,而且37lines实现的，受教了！
递归的时候模拟一个栈，把树按根，右子树，左子树存入，再LIFO输出
代码
#include
<
stdio.h
>
#define
M 1001
int
stack[M];
int
top;
void
Create(
int
*
pre,
int
plen,
int
*
in
,
int
ilen)
{
int
index;
if
(plen)
{
stack[top
++
]
=
pre[
0
];
for
(index
=
0
;
in
[index]
!=
pre[
0
];index
++
);
Create(pre
+
index
+
1
,plen
-
index
-
1
,
in
+
index
+
1
,ilen
-
index
-
1
);
Create(pre
+
1
,index,
in
,index);
}
}
int
main()
{
int
n,i;
int
pre[M],
in
[M];
while
(scanf(
"
%d
"
,
&
n)
!=-
1
){
for
(i
=
0
;i
<
n;i
++
)
scanf(
"
%d
"
,pre
+
i);
for
(i
=
0
;i
<
n;i
++
)
scanf(
"
%d
"
,
in
+
i);
top
=
0
;
Create(pre,n,
in
,n);
while
(top
--
)
{
if
(top
==
0
){
printf(
"
%d\n
"
,stack[top]);
break
;
}
printf(
"
%d
"
,stack[top]);
}
}
return
0
;
}