二叉树的Morris遍历

此算法可以时间复杂度O(N),空间复杂度O（1）的遍历二叉树

算法流程：

1.首先cur指向根节点，判断是否有左子树，如果没有左子树则cur指向右子树

2.若有左子树，则找到其左子树的最右边的节点，判断此节点的右树是否指向cur节点（判断是第一次到还是第二次到），若指向空，将此节点的右树指向cur，并将cur移向左子树。若指向cur,则恢复成指向空。并将cur移向右子树

3.重复执行1到2步骤，直到cur指向空。

 

vector<char> Morris(TreeNode *root)
{
	vector<char> res;
	if (root == NULL)
		return{};
	TreeNode *cur = root;
	TreeNode *MostRight = NULL;
	while (cur != NULL)
	{
		res.push_back(cur->val);
		if (cur->left != NULL)//会来到两次的节点
		{
			MostRight = cur->left;
			while (MostRight->right != NULL&&MostRight->right!=cur)
			{
				MostRight = MostRight->right;
			}
			if (MostRight->right == NULL)//第一次来到此节点
			{
				MostRight->right = cur;
				cur = cur->left;
			}
			else//第二次来到此节点
			{
				MostRight->right = NULL;
				cur = cur->right;
			}
		}
		else//只可能来一次的节点
		{
			cur = cur->right;
		}
	}
	return res;
}

此流程会输出Morris序，有左子树的节点都会访问两次，其他节点一次。

若要实现先序遍历，则在每次第一次访问节点时输出即可。

#include<iostream>
#include<vector>

using namespace std;

struct TreeNode
{
	char val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int a):val(a),left(NULL),right(NULL){}
};
TreeNode *CreateTree(string s, int &index);
TreeNode *Buy_Node(int val);
vector<char> Morris(TreeNode *root);
int main()
{
	string s = { "ABC#D#EF####GH##I#J##" };
	int index = 0;
	TreeNode *root = CreateTree(s, index);
	vector<char> res = Morris(root);
}

TreeNode *CreateTree(string s,int &index)
{
	if (s[index] == '#')
		return NULL;
	TreeNode *root = Buy_Node(s[index]);
	root->left = CreateTree(s, ++index);
	root->right = CreateTree(s, ++index);
	return root;
}

TreeNode *Buy_Node(int val)
{
	return new TreeNode(val);
}

vector<char> Morris(TreeNode *root)
{
	vector<char> res;
	if (root == NULL)
		return{};
	TreeNode *cur = root;
	TreeNode *MostRight = NULL;
	while (cur != NULL)
	{
		res.push_back(cur->val);
		if (cur->left != NULL)//会来到两次的节点
		{
			MostRight = cur->left;
			while (MostRight->right != NULL&&MostRight->right!=cur)
			{
				MostRight = MostRight->right;
			}
			if (MostRight->right == NULL)//第一次来到此节点
			{
				cout << cur->val;
				MostRight->right = cur;
				cur = cur->left;
			}
			else//第二次来到此节点
			{
				MostRight->right = NULL;
				cur = cur->right;
			}
		}
		else//只可能来一次的节点
		{
			cout << cur->val;
			cur = cur->right;
		}
	}
	return res;
}

若想中序遍历，则遇到需要访问两次的节点，第二次访问的时候输出，只会访问一次的节点，直接输出。

#include<iostream>
#include<vector>

using namespace std;

struct TreeNode
{
	char val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int a):val(a),left(NULL),right(NULL){}
};
TreeNode *CreateTree(string s, int &index);
TreeNode *Buy_Node(int val);
vector<char> Morris(TreeNode *root);
int main()
{
	string s = { "ABC#D#EF####GH##I#J##" };
	int index = 0;
	TreeNode *root = CreateTree(s, index);
	vector<char> res = Morris(root);
}

TreeNode *CreateTree(string s,int &index)
{
	if (s[index] == '#')
		return NULL;
	TreeNode *root = Buy_Node(s[index]);
	root->left = CreateTree(s, ++index);
	root->right = CreateTree(s, ++index);
	return root;
}

TreeNode *Buy_Node(int val)
{
	return new TreeNode(val);
}

vector<char> Morris(TreeNode *root)
{
	vector<char> res;
	if (root == NULL)
		return{};
	TreeNode *cur = root;
	TreeNode *MostRight = NULL;
	while (cur != NULL)
	{
		res.push_back(cur->val);
		if (cur->left != NULL)//会来到两次的节点
		{
			MostRight = cur->left;
			while (MostRight->right != NULL&&MostRight->right!=cur)
			{
				MostRight = MostRight->right;
			}
			if (MostRight->right == NULL)//第一次来到此节点
			{
				MostRight->right = cur;
				cur = cur->left;
			}
			else//第二次来到此节点
			{
				cout << cur->val;
				MostRight->right = NULL;
				cur = cur->right;
			}
		}
		else//只可能来一次的节点
		{
			cout << cur->val;
			cur = cur->right;
		}
	}
	return res;
}

若想后序遍历则比较复杂，每次第二次访问节点时，就逆序输出此节点左节点的右边界。最后再逆序输出整个树的右边界

vector<char> Morris(TreeNode *root)
{
	vector<char> res;
	if (root == NULL)
		return{};
	TreeNode *cur = root;
	TreeNode *MostRight = NULL;
	while (cur != NULL)
	{
		res.push_back(cur->val);
		if (cur->left != NULL)//会来到两次的节点
		{
			MostRight = cur->left;
			while (MostRight->right != NULL&&MostRight->right!=cur)
			{
				MostRight = MostRight->right;
			}
			if (MostRight->right == NULL)//第一次来到此节点
			{
				MostRight->right = cur;
				cur = cur->left;
			}
			else//第二次来到此节点
			{
				MostRight->right = NULL;
				PrintRightEdge(cur->left);
				cur = cur->right;
			}
		}
		else//只可能来一次的节点
		{
			cur = cur->right;
		}
	}
	PrintRightEdge(root);
	return res;
}

TreeNode *reverse(TreeNode *head)
{
	TreeNode *pre = NULL;
	TreeNode *cur = head;
	while (cur != NULL)
	{
		TreeNode *next = cur->right;
		cur->right = pre;
		pre = cur;
		cur = next;
	}
	return pre;
}

void PrintRightEdge(TreeNode *head)
{
	TreeNode *tail = reverse(head);
	TreeNode *p = tail;
	while (p!= NULL)
	{
		cout << p->val;
		p = p->right;
	}
	reverse(tail);
}