三种方法中，递归最为简单，栈次之，循环最为麻烦。递归的深度如果太大则会导致栈溢出；栈的方式需要额外的辅助空间；循环编程最麻烦。
首先是递归：
//递归方法
void midPrint_r(TreeNode* root)
{//中序遍历
if(root==NULL)
return;
if(root->left)
midPrint(root->left);
cout<<root->val<<" ";
if(root->right)
midPrint(root->right);
}
void prePrint_r(TreeNode* root)
{//前序遍历
if(root==NULL)
return;
cout<<root->val<<" ";
if(root->left)
prePrint(root->left);
if(root->right)
prePrint(root->right);
}
void postPrint_r(TreeNode* root)
{//中序遍历
if(root==NULL)
return;
if(root->left)
postPrint(root->left);
if(root->right)
postPrint(root->right);
cout<<root->val<<" ";
}//循环堆栈方法
void midPrint_l(TreeNode* root)
{
stack<TreeNode*> s;
TreeNode *cur = root;
while(!s.empty() || cur != NULL)
{
while(cur != NULL)
{
s.push(cur);
cur = cur->left;
}
cur = s.top();
s.pop();
cout<<cur->val<<" ";
cur = cur->right;
}
}
void prePrint_l(TreeNode* root)
{
stack<TreeNode*> s;
TreeNode *cur = root;
while(!s.empty() || cur != NULL)
{
while(cur != NULL)
{
cout<<cur->label<<" ";
s.push(cur);
cur = cur->left;
}
cur = s.top();
s.pop();
cur = cur->right;
}
}
void postPrint_l(TreeNode* root)
{
//the original method
/*
We use a prev variable to keep track of the previously-traversed node.
Let’s assume curr is the current node that’s on top of the stack. When
prev is curr‘s parent, we are traversing down the tree. In this case,
we try to traverse to curr‘s left child if available (ie, push left
child to the stack). If it is not available, we look at curr‘s right
child. If both left and right child do not exist (ie, curr is a leaf node),
we print curr‘s value and pop it off the stack.If prev is curr‘s left
child, we are traversing up the tree from the left. We look at curr‘s
right child. If it is available, then traverse down the right child (ie,
push right child to the stack), otherwise print curr‘s value and pop it
off the stack.If prev is curr‘s right child, we are traversing up the tree
from the right. In this case, we print curr‘s value and pop it off the stack.
*/
if(root == NULL)
{
return;
}
stack<TreeNode*> s;
s.push(root);
TreeNode *cur = NULL;
TreeNode *pre = NULL;
while(!s.empty())
{
cur = s.top();
//left down or right down
if( pre == NULL || pre->left == cur || pre->right == cur)
{//有左子树压左子树，有右子树压右子树，
//都没有说明是叶子结点，直接打印并弹出
if(cur->left != NULL)
{//先压左子树，若左子树为空则压右子树
s.push(cur->left);
}
else if(cur->right != NULL)
{
s.push(cur->right);
}
else
{//左右子树均为空
cout<<cur->label<<" ";
s.pop();
}
}
//left up
else if(cur->left == pre)
{//左边开始返回
if(cur->right != NULL)
{//若右孩子不为空则压入，否则，打印
s.push(cur->right);
}
else
{
cout<<cur->label<<" ";
s.pop();
}
}
//right up
else if(cur->right == pre)
{//从右边返回,说明右侧已经打印完，则可以直接打印
cout<<cur->label<<" ";
s.pop();
}
pre = cur;
}
// the easy method
/*stack<TreeNode*> s;
s.push(root);
TreeNode *cur = NULL;
TreeNode *pre = NULL;
while(!s.empty())
{
cur = s.top();
if(pre == NULL || pre->left == cur || pre->right == cur)
{
if(cur->left != NULL)
{
s.push(cur->left);
}
else if(cur->right != NULL)
{
s.push(cur->right);
}
}
else if(cur->left == pre)
{
if(cur->right != NULL)
{
s.push(cur->right);
}
}
else
{
cout<<cur->label<<" ";
s.pop();
}
pre = cur;
}*/
}
//不用辅助空间的方式
void midPrint_m(TreeNode *root)
{
TreeNode *cur = root;
TreeNode *pre = NULL;
while(cur != NULL)
{
if(cur->left == NULL)
{//左孩子为空，则直接输出
cout<<cur->label<<" ";
cur = cur->right;
}
else
{
pre = cur->left;
while( pre->right != NULL && pre->right != cur )
{//找到cur的直接前驱
pre = pre->right;
}
if(pre->right == NULL)
{//设置后继结点
pre->right = cur;
cur = cur->left;
}
else
{
pre->right = NULL;//重新设置为空
cout<<cur->label<<" ";
cur = cur->right;
}
}
}
}
void prePrint_m(TreeNode *root)
{//基本同于中遍历
TreeNode *cur = root;
TreeNode *pre = NULL;
while(cur != NULL)
{
if(cur->left == NULL)
{
cout<<cur->label<<" ";
cur = cur->right;
}
else
{
pre = cur->left;
while( pre->right != NULL && pre->right != cur )
{
pre = pre->right;
}
if(pre->right == NULL)
{
pre->right = cur;
cout<<cur->label<<" ";
cur = cur->left;
}
else
{
pre->right = NULL;
cur = cur->right;
}
}
}
}
//this is the most difficult algorithm
void reverse_out(TreeNode *from,TreeNode *to)
{
//first reverse from->to
//reverse
TreeNode *cur = from;
TreeNode *post = cur->right;
while(cur != to)
{
TreeNode *tmp = post->right;
post->right = cur;
cur = post;
post = tmp;
}
//already reverse,output list
TreeNode *traversal = cur;
while( cur != from )
{
cout<<cur->label<<" ";
cur = cur->right;
}
cout<<cur->label<<" ";
//reverse original
cur = to;
post = cur->right;
while(cur != from)
{
TreeNode *tmp = post->right;
post->right = cur;
cur = post;
post = tmp;
}
//restore to's right
to->right = NULL;
}
void postPrint_m(TreeNode *root)
{
TreeNode *newroot = new TreeNode(0);
newroot->left = root;
newroot->right = NULL;
TreeNode *cur = newroot;
TreeNode *pre = NULL;
while(cur != NULL)
{
if(cur->left == NULL)
{
cur = cur->right;
}
else
{
pre = cur->left;
while(pre->right != NULL && pre->right != cur)
{
pre = pre->right;
}
if(pre->right == NULL)
{
pre->right = cur;
cur = cur->left;
}
else
{
pre->right = NULL;
reverse_out(cur->left,pre);
cur = cur->right;
}
}
}
}原文地址：http://blog.csdn.net/walker19900515/article/details/46580527
