［算法］二叉树的遍历：前序，中序与后序

二叉树的遍历是二叉树的众多算法的基础。主要有，前序，中序与后序。

对于以下二叉树：

前序：12354

中序：21543

后序：24531

笔者实现了三种遍历方式:

1  前序：递归版本比较简单，只需要改变push_back操作的位置即可。

vector PreOrderTraverse2(TreeNode *root)
{
	vector nodes;
	vector temp;
	if(root == NULL) return nodes;
	nodes.push_back(root->val);
	temp = PreOrderTraverse2(root->left);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	temp = PreOrderTraverse2(root->right);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	return nodes;

}

 前序迭代版本，步骤如下：

（1）首先访问根节点root，先将root节点输出，压入堆栈，

（2）再考虑root节点的左子树，左子树也是一颗二叉树，重复（1）。

（3）某一步的左子树为空，那么该节点左子树已访问完毕，弹出栈顶，访问该节点的右子树，重复（1）即可。

很简单，只需要记住，在将节点压入堆栈时，该节点就已被访问，再访问完左孩子之后，再考虑右孩子所在的二叉树。

vector PreOrderTraverse(TreeNode *root)
{
	vector nodes;
	stack s;
	if(root == NULL) return nodes;
	TreeNode *temp = root;
	while(temp || !s.empty())
	{
		while(temp)
		{
			s.push(temp);//入栈之前就访问
			nodes.push_back(temp->val);
			temp = temp->left;
		}//左子树访问完毕
		temp = s.top();
		s.pop();
		temp = temp->right;
	}
	return nodes;
}

2  中序， 递归版本：

vector InOrderTraverse2(TreeNode *root)
{
	vector nodes;
	vector temp;
	if(root == NULL) return nodes;
	temp = InOrderTraverse2(root->left);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	nodes.push_back(root->val);
	temp = InOrderTraverse2(root->right);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	return nodes;

}

迭代版本：

（1）首先访问根节点root，压入堆栈。

（2）访问root的左子树，重复（1）即可。

（3）如果某节点左子树为空，那么该节点输出，并考察其右子树，重复（1）即可。

PS：节点入栈使不访问，遇到节点就入栈，直到栈顶左子树访问完毕，那么将栈顶弹出，考虑右子树，重复即可。

vector InOrderTraverse(TreeNode *root)
{
	vector nodes;
	if(root == NULL) return nodes;
	TreeNode *temp = root;
	stack s;

	while(temp || !s.empty())
	{
		while(temp)
		{
			s.push(temp);
			temp = temp->left;
		}// 左子树完毕
		temp = s.top();
		nodes.push_back(temp->val);
		s.pop();
		temp = temp->right;
	}
	return nodes;
}

3 后序，递归版本：

vector PostOrderTraverse2(TreeNode *root)
{
	vector nodes;
	vector temp;
	if(root == NULL) return nodes;
	temp = PostOrderTraverse2(root->left);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	temp = PostOrderTraverse2(root->right);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	nodes.push_back(root->val);
	return nodes;

}

迭代版本：步骤如下：

PS：遇到根节点root，分两种情况讨论，如果根节点的右子树还没被访问，那么根节点入栈；否则输出根节点。

vector PostOrderTraverse(TreeNode *root)
{
	vector nodes;
	if(root == NULL) return nodes;
	TreeNode *temp = root;
	TreeNode *pre = root; //记录前一次访问的节点
	stack s;
	while(temp || !s.empty())
	{
		while(temp)
		{
			s.push(temp);
			temp = temp->left;
		}
		temp = s.top();
		if (temp->right == NULL || pre == temp->right)
		{
			nodes.push_back(temp->val);
			pre = temp;
			s.pop();
			temp = NULL;//保证不重复访问
		}
		else
		{
			temp = temp->right;
		}
	}
	return nodes;
}

后序遍历的递归版本比较复杂。自己在编写代码时会出现死循环，因为出现了重复访问，因此上述的temp=NULL语句很重要。

后记：上述3种算法的空间复杂度均为O（n）。

完整的代码为：

#include
#include
#include

using namespace std;

struct TreeNode{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x):val(x),left(NULL),right(NULL){}
};

vector InOrderTraverse2(TreeNode *root)
{
	vector nodes;
	vector temp;
	if(root == NULL) return nodes;
	temp = InOrderTraverse2(root->left);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	nodes.push_back(root->val);
	temp = InOrderTraverse2(root->right);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	return nodes;

}

vector PreOrderTraverse2(TreeNode *root)
{
	vector nodes;
	vector temp;
	if(root == NULL) return nodes;
	nodes.push_back(root->val);
	temp = PreOrderTraverse2(root->left);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	temp = PreOrderTraverse2(root->right);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	return nodes;

}


vector PostOrderTraverse2(TreeNode *root)
{
	vector nodes;
	vector temp;
	if(root == NULL) return nodes;
	temp = PostOrderTraverse2(root->left);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	temp = PostOrderTraverse2(root->right);
	for(vector::size_type i = 0 ; i != temp.size() ; ++i)
		nodes.push_back(temp[i]);
	nodes.push_back(root->val);
	return nodes;

}

vector PreOrderTraverse(TreeNode *root)
{
	vector nodes;
	stack s;
	if(root == NULL) return nodes;
	TreeNode *temp = root;
	while(temp || !s.empty())
	{
		while(temp)
		{
			s.push(temp);//入栈之前就访问
			nodes.push_back(temp->val);
			temp = temp->left;
		}
		temp = s.top();
		s.pop();
		temp = temp->right;
	}
	return nodes;
}

vector InOrderTraverse(TreeNode *root)
{
	vector nodes;
	if(root == NULL) return nodes;
	TreeNode *temp = root;
	stack s;

	while(temp || !s.empty())
	{
		while(temp)
		{
			s.push(temp);
			temp = temp->left;
		}
		temp = s.top();
		nodes.push_back(temp->val);
		s.pop();
		temp = temp->right;
	}
	return nodes;
}

vector PostOrderTraverse(TreeNode *root)
{
	vector nodes;
	if(root == NULL) return nodes;
	TreeNode *temp = root;
	TreeNode *pre = root; //记录前一次访问的节点
	stack s;
	while(temp || !s.empty())
	{
		while(temp)
		{
			s.push(temp);
			temp = temp->left;
		}
		temp = s.top();
		if (temp->right == NULL || pre == temp->right)
		{
			nodes.push_back(temp->val);
			pre = temp;
			s.pop();
			temp = NULL;//保证不重复访问
		}
		else
		{
			temp = temp->right;
		}
	}
	return nodes;
}



int main()
{
	vector nodes;
	TreeNode *root = new TreeNode(1);
	TreeNode *node2 = new TreeNode(2);
	TreeNode *node3 = new TreeNode(3);
	TreeNode *node4 = new TreeNode(4);
	TreeNode *node5 = new TreeNode(5);
	root->left = node2;
	root->right = node3;
	node5->right = node4;
	node3->left = node5;
	
	cout << "InOrder recursively..." << endl;
	nodes = InOrderTraverse2(root);
	for(vector::size_type i = 0; i != nodes.size(); i++)
		cout << nodes[i] << " ";
	cout << endl;
	cout << "InOrder iterativelly..." << endl;
	nodes = InOrderTraverse(root);
	for(vector::size_type i = 0; i != nodes.size(); i++)
		cout << nodes[i] << " ";
	cout << endl;
	cout << "PreOrder recursively..." << endl;
	nodes = PreOrderTraverse2(root);
	for(vector::size_type i = 0; i != nodes.size(); i++)
		cout << nodes[i] << " ";
	cout << endl;
	cout << "PreOrder iterativelly..." << endl;
	nodes = PreOrderTraverse(root);
	for(vector::size_type i = 0; i != nodes.size(); i++)
		cout << nodes[i] << " ";
	cout << endl;
	cout << "PostOrder recursively..." << endl;
	nodes = PostOrderTraverse2(root);
	for(vector::size_type i = 0; i != nodes.size(); i++)
		cout << nodes[i] << " ";
	cout << endl;
	cout << "PostOrder iterativelly..." << endl;
	nodes = PostOrderTraverse(root);
	for(vector::size_type i = 0; i != nodes.size(); i++)
		cout << nodes[i] << " ";
	cout << endl;
}