二叉树的morris、层序遍历

Morris遍历
是一种时间复杂度为O(N)，空间复杂度为O(1)的遍历方式
可以利用morris遍历实现二叉树的先序、中序、后序遍历

流程说明：
当前节点为cur
①cur无左子树， cur = cur.right;
②cur有右子树，找到左子树上最右节点mostright
1°if mostright.right == null，mostright.right = cur， cur = cur.left；
2°if mostright.right == cur， mostright.right = null， cur = cur.right；
③ ②中的1° 2°分别表示第一次遍历到该节点和第二次遍历到该节点，若第一次遍历到该节点就打印的话就是先序遍历，第二次遇到再打印属于中序遍历

morris的先序遍历

void morrispre(node head){
	if(head == NULL)
		return;
	node cur1 = head, cur2 = null;
	while(cur1 != NULL){
		cur2 = cur1.left;
		if(cur2 != NULL){
			while(cur2.right != NULL && cur2.right != cur1)
				cur2 = cur2.right;
			if(cur2.right == NULL){
				cur2.right = cur1;
				cout << cur1.data << endl;
				cur1 = cur1.left;
				continue;
			}
			else{
				cur2.right = NULL;
				//cout << cur1.data << endl;
				cur1 = cur1.right;
			}
		}else{
			cout << cur1.data << endl;
			cur1 = cur1.right;
		}
	}
}

morris的中序遍历

void morrispre(node head){
	if(head == NULL)
		return;
	node cur1 = head, cur2 = null;
	while(cur1 != NULL){
		cur2 = cur1.left;
		if(cur2 != NULL){
			while(cur2.right != NULL && cur2.right != cur1)
				cur2 = cur2.right;
			if(cur2.right == NULL){
				cur2.right = cur1;
				//cout << cur1.data << endl;
				cur1 = cur1.left;
				continue;
			}
			else{
				cur2.right = NULL;
				cout << cur1.data << endl;
				cur1 = cur1.right;
			}
		}else{
			//cout << cur1.data << endl;
			cur1 = cur1.right;
		}
	}
}

morris的后序遍历
后序遍历就不是更改打印位置就可以的了，而是当第二次遇到该节点时，就逆序打印该节点左子树的最右层，最后打印整棵树的最右层

void morrispre(node head){
	if(head == NULL)
		return;
	node cur1 = head, cur2 = null;
	while(cur1 != NULL){
		cur2 = cur1.left;
		if(cur2 != NULL){
			while(cur2.right != NULL && cur2.right != cur1)
				cur2 = cur2.right;
			if(cur2.right == NULL){
				cur2.right = cur1;
				//cout << cur1.data << endl;
				cur1 = cur1.left;
				continue;
			}
			else{
				cur2.right = NULL;
				//cout << cur1.data << endl;
				printfEdge(cur1.left);
				cur1 = cur1.right;
			}
		}else{
			//cout << cur1.data << endl;
			cur1 = cur1.right;
		}
	}
	printfEdge(head);
}

【层序遍历】
类似图的广度优先搜索
基本过程：
①先把根节点放入队列
②进入while循环，条件是队列不为空
1°从队列中取出一个元素， 并打印
2°若该节点左右孩子不为空，则顺序入队
③while循环结束

伪代码

void LevelOrderTraversal（BinTree BT）{
	Queue Q; BinTree T;
	Add(Q, BT);
	while(!isempty(Q)){
		T = Delete(Q);
		if(T->Left)	AddQ(Q, T->Left);
		if(T->Right)	AddQ(Q, T->Right);
	}
}