非递归方法实现 二叉树的 DFS(Pre Order), Post Order

Post Order比较难以理解，开辟的lastVisited记录上次访问的点，如果这个点是当前点的右孩子，则打印当前点 

#include <iostream>
#include <stack>


struct node
{
	int data;
	node* left;
	node* right;
	node(int exData)
	{
		data = exData;
		left = NULL;
		right = NULL;
	}
};

void dfs(node* root)
{
	std::stack<node*> st;
	node* cur = root;

	while(cur || !st.empty())
	{
		while(cur)
		{
			st.push(cur);
			std::cout << cur->data << " ";
			cur = cur->left;
		}

		if(!st.empty())
		{
			cur = st.top();
			cur = cur->right;
			st.pop();
		}
	}
};

void postOrderTrav(node *root)
{
	std::stack<node*> st;
	node* cur = root;
	node* lastVisited = NULL;
	node* p = NULL;
	while(cur || !st.empty())
	{
		while(cur)
		{
			st.push(cur);
			cur = cur->left;
		}
		p = st.top();
		if(!p->right || p->right == lastVisited)
		{
			std::cout << p->data << " ";
			lastVisited = p;
			st.pop();
		}
		else
		{
			cur = p