树的遍历

递归：

1.先序

void PreOrder(T)

{

if T !=  NULL

Print T

PreOrder(T.lchild)

PreOrder(T.rchild)

}

2.中序

void InOrder(T)

{

if T !=  NULL

InOrder(T.lchild)

Print T

InOrder(T.rchild)

}

3.后序

void PostOrder(T)

{

if T !=  NULL

PostOrder(T.lchild)

PostOrder(T.rchild)

Print T

}

非递归：

1.先序：

void PreOrder(T)

{

t = T

Stack stack

while t != NULL || !stack.empty()

if t != NULL

print t

Push(stack, t)

t = t.lchild

else

t = Pop(stack)

t = t.rchild

}

2.中序

void InOrder(T)

{

t = T

Stack stack

while t != NULL || !stack.empty()

if t != NULL

Push(stack, t)

t = tlchild

else

t = Pop()

print t

t = t.rchild

}

3.后序

后序遍历由于遍历父节点是在遍历子节点之后，而且左节点和右节点遍历后的行为不一样，

因此需要用变量来记录前一次访问的节点，根据前一次节点和现节点的关系确定操作。http://blog.csdn.net/hackbuteer1/article/details/6583988

void PostOrder(T)

{

if  T == NULL

return;

Stack stack

Push(stack, T)

while(!stack.empty())

{

cur = s.top()

if (pre == NULL || pre->lchild == cur || pre->rchild == cur)

{

if (cur->lchild != NULL)

Push(stack, cur->lchild)

else if (cur->rchild != NULL)

Push(stack, cur->rchild)

}

else if (cur->lchild == pre)

{

if (cur->rchild != NULL)

Push(stack, cur->rchild)

}

else

{

print cur

s.pop()

}

pre = cur

}

}

不用栈非递归中序遍历(需要父节点）while(t != null){    while(t->left && p != t->left && p != t->right)    {<span style="white-space:pre"></span>p = t;<span style="white-space:pre"></span>t = t->left;    }    if(p != t->right)    {    <span style="white-space:pre"></span>print t;<span style="white-space:pre"></span>if(t->right)<span style="white-space:pre"></span>{<span style="white-space:pre"></span>    p = t;<span style="white-space:pre"></span>    t = t->right;<span style="white-space:pre"></span>}    }    else    {<span style="white-space:pre"></span>p = t;<span style="white-space:pre"></span>t = t->parent;    }    }