二叉树的前中后序非递归遍历

// 前序和中序比较简单，不做解释，后序遍历做了一定解释
#include <iostream>
#include <stack>
using namespace std;
struct TreeNode 
{
    int m_data;
    TreeNode * leftChild;
    TreeNode * rightChild;
    TreeNode(int data) : m_data(data), leftChild(NULL), rightChild(NULL) {}
};

void preOrderTravel(TreeNode * root)
{
    stack< TreeNode * > st;
    while( !st.empty() || root != NULL )
    {
        while( root )
        {
            st.push( root );
            cout << root->m_data << " ";
            root = root->leftChild;
        }
        root = st.top();
        st.pop();
        root = root->rightChild;
    }
}
void inOrderTravel(TreeNode * root)
{
     stack< TreeNode * > st;
     while( !st.empty() || root != NULL )
     {
         while( root )
         {
             st.push( root );
             root = root->leftChild;
         }
         root = st.top();
         st.pop();
         cout << root->m_data << " ";
         root = root->rightChild;
    }
}
// 后序遍历由于其有一定复杂性，需要构造一个辅助节点来
// 帮助，此节点会多2个标记，标记该结点是否已经遍历了左
// 右子树，可以输出该节点
struct Node 
{
    TreeNode * m_node;
    int leftTag;
    int rightTag;

    // 要是该节点有左结点，则该节点标记置1，否则置0
    Node(TreeNode * root)
    {
         m_node = root;
         leftTag = root->leftChild ? 1 : 0;
         rightTag = root->rightChild ? 1 : 0;
    }
    // 重载类型转换运算符，只是为了操作简便
    operator TreeNode* ()
    {
         return m_node;
    }
    // 重载->运算符，只是为了操作简便
    TreeNode * operator->()
    {
         return m_node;
    }
};
void postOrderTravel(TreeNode * root)
{
    if( root == NULL ) return;
    Node node = root;
    stack< Node > stn;
    while( true )
    {
         // 若此节点有左子树，且左子树尚未被遍历，则遍历
         // 这里不需要考虑 node 节点为空的情况，因为叶子
         // 节点的 leftTag/rightTag 已经置为0啦
         while( node.leftTag )
         {
             // 一旦遍历则直接令其为0，表示已经遍历
             node.leftTag = 0;
             stn.push( node );
             node = node->leftChild;
         }
         // 进入这一步，表示左节点已经到了最左，只需判断该
         // 节点是否有有右节点，因为这是后序遍历，根节点一定
         // 在左右遍历完之后才输出
         if( node.rightTag )
         {
             node.rightTag = 0;
             stn.push( node );
             node = node->rightChild;
         }
         else 
         {
             cout << node->m_data << " ";
             // 当最后一个节点输出之后，这里就可跳出
             if( stn.empty() ) break;
             node = stn.top();
             stn.pop();
         }
    }
}
int main()
{
     // 测试数据
     TreeNode * root  = new TreeNode( 1 );
     TreeNode * node1 = new TreeNode( 2 );
     TreeNode * node2 = new TreeNode( 3 );
     TreeNode * node3 = new TreeNode( 4 );
     TreeNode * node4 = new TreeNode( 5 );
     TreeNode * node5 = new TreeNode( 6 );
     TreeNode * node6 = new TreeNode( 7 );
     TreeNode * node7 = new TreeNode( 8 );
     root->leftChild   = node1;
     root->rightChild  = node4;
     node1->leftChild  = node2;
     node1->rightChild = node3;
     node4->leftChild  = node5;
     node4->rightChild = node6;
     node6->rightChild = node7;
     cout << "前序遍历结果：" << endl;
     preOrderTravel( root );
     cout << endl;
     cout << "中序遍历结果：" << endl;
     inOrderTravel( root );
     cout << endl;
     cout << "后序遍历结果：" << endl;
     postOrderTravel( root );
     cout << endl;
     return 0;
}