Four binary tree traversal algorithm

Now there is such a binary tree:

Preorder

Ideas and Code

According to: node, the left sub-tree, the order of traversal right subtree

void PriorOrder(BTNode* b) 
{
	if(b)
	{
		printf("%c", b->data);
		PriorOrder(b->lchild);
		PriorOrder(b->rchild);	
	}
} 

Process Analysis

At first, we will first traverse the topmost tree (the red box)
in three parts this number: the root, left subtree, right subtree, respectively, circled by a blue box
we first access node A and then visit the left subtree, which is on the left in the blue box as a whole
a

Now separate the left sub-tree made of view, is the same: first visit B, then visit the left node, D, and then visit the right subtree.

We then raised the right subtree: still the same, then access to access E H, then and now, the beginning and the left sub-tree root node consisting of a number of ABC have visited over, so now begin to access the right node number of the ABC

ABC right node of the tree:

the same as above, can be accessed.
The final access order: ABDEHCFIJG

Preorder

Ideas and Code

According to: order left node, this node, right node traversal

void InOrder(BTNode* b) 
{
	if(b)
	{
		InOrder(b->lchild);
		printf("%c", b->data);
		InOrder(b->rchild);	
	}
} 

The process is still the same as above, except that the left subtree first visit complete.
Output order: DBHEAFJICG

Subsequent traversal

Ideas and Code

According to: order the left node, right node, this node traversal

void PostOrder(BTNode* b) 
{
	if(b)
	{
		PostOrder(b->lchild);
		PostOrder(b->rchild);
		printf("%c", b->data);	
	}
} 

Output order: DHEBJIFGCA

Traverse the level

Ideas and Code

According to: root, the order of the first layer, the second layer traversal ...

void LevelTraversal(BTNode* b)
{
	if(b)
	{
		//定义队列 
		BTNode* queue[MaxSize];
		int top=0;
		queue[top] = b;
		int i = 0;
		
		//只要队列里有元素 
		while(i<=top)
		{
			//将左右子树加入队列 
			if(queue[i]->lchild)
			{
				top++;
				queue[top] = queue[i]->lchild;	
			}
			if(queue[i]->rchild)
			{
				top++;
				queue[top] = queue[i]->rchild;	
			}
			i++;
		}

		//现在这个队列按层次顺序容纳了二叉树的每个节点	
		for(i=0;i<=top;i++)
		{
			//挨个打出来即可 
			printf("%c", queue[i]->data);
		}			
		//我比较懒所以就没写出队，这其实当成个栈也可以 
	}  			
} 

It is clear that is one level of access, the order is ABCDEFGHIJ