二叉树的层序遍历

二叉树层序遍历及优化算法

最新推荐文章于 2024-03-25 19:51:54 发布

翻译最新推荐文章于 2024-03-25 19:51:54 发布 · 4.4k 阅读

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本文介绍了一种二叉树层序遍历的实现方式，并通过STL库优化了代码效率。同时提供了拓展问题及解决方案，包括访问顺序调整、使用栈和队列进行遍历等。

/* * Copyright (c) 2011 alexingcool. All Rights Reserved. */ #include <iostream> using namespace std; struct Node { Node(int i = 0, Node *l = NULL, Node *r = NULL) : data(i), left(l), right(r) {} int data; Node *left; Node *right; }; Node* Construct() { Node *node4 = new Node(7, NULL, new Node(3)); Node *node3 = new Node(4); Node *node2 = new Node(12); Node *node1 = new Node(5, node3, node4); Node *root = new Node(10, node1, node2); return root; } int PrintByLevel(Node *root, int level) { if(root == NULL) return 0; if(level == 1) { cout << root->data << " "; return 1; } return PrintByLevel(root->left, level - 1) + PrintByLevel(root->right, level - 1); } int GetTreeHeight(Node *root) { if(root == NULL) return 0; return max(GetTreeHeight(root->left) + 1, GetTreeHeight(root->right) + 1); } void main() { Node *root = Construct(); int height = GetTreeHeight(root); for(int i = 1; i <= height; i++) { PrintByLevel(root, i); cout << endl; } }

关于层序遍历的STL代码如下，注意reserve的妙用，否则会因为vector容量增长导致跌代器失效：

void Print2(Node *head) { vector<Node*> nvec; nvec.reserve(100); nvec.push_back(head); while(!nvec.empty()) { int size = nvec.size(); for(vector<Node*>::const_iterator iter = nvec.begin(); iter != nvec.end(); ++iter) { cout << (*iter)->data << " "; } cout << endl; for(vector<Node*>::const_iterator iter = nvec.begin(); iter != nvec.begin() + size; ++iter) { if((*iter)->left) nvec.push_back((*iter)->left); if((*iter)->right) nvec.push_back((*iter)->right); } nvec.erase(nvec.begin(), nvec.begin() + size); } }

下面代码不知为何总报错，伤感了，眼看找工作高峰来临，时间紧迫，扔这了。。。

void printLevel(Node *root)
{
	if(root == NULL)
		return;

	vector<Node *> nvec;
	nvec.reserve(100);
	nvec.push_back(root);
	vector<Node *>::iterator startIter = nvec.begin();
	vector<Node *>::iterator lastIter = nvec.end();

	while(startIter != lastIter) {
		while(startIter != lastIter) {
			cout << (*startIter)->data << " ";
			Node *left = (*startIter)->left;
			Node *right = (*startIter)->right;
			if(left)
				nvec.push_back(left);
			if(right)
				nvec.push_back(right);
			++startIter;
		}
		cout << endl;
		lastIter = nvec.end();
	}
}

再次写二叉树的层序遍历，现在的代码可以说是简洁高效，窝稀饭

void printLevel(Node *root)
{
	if (root == NULL)
		return;

	vector<Node *> nvec;
	nvec.push_back(root);
	int start = 0;
	int end = 1;
	int pos = end;

	while (start != end)
	{
		Node *node = nvec[start];
		cout << node->item << " ";
		start++;

		if (node->left)
		{
			nvec.push_back(node->left);
			end++;
		}
		if (node->right)
		{
			nvec.push_back(node->right);
			end++;
		}

		if (start == pos)
		{
			pos = end;
			cout << endl;
		}
	}
}

关于拓展问题：

访问顺序变为：

456

访问顺序变为：

654

用栈很容易实现~

void putNode2Stack(Node *root, stack<Node *> &snode)
{
	if (root == NULL)
		return;

	vector<Node *> nvec;
	nvec.push_back(root);
	snode.push(root);
	snode.push(NULL);
	int start = 0;
	int end = 1;
	int pos = end;

	while (start != end)
	{
		Node *node = nvec[start];
		cout << node->item << " ";
		start++;

		if (node->left)
		{
			nvec.push_back(node->left);
			snode.push(node->left);
			end++;
		}
		if (node->right)
		{
			nvec.push_back(node->right);
			snode.push(node->right);
			end++;
		}

		if (start == pos)
		{
			pos = end;
			snode.push(NULL);
			cout << endl;
		}
	}
}

void putNode2Stack2(Node *root, stack<Node *> &snode)
{
	if (root == NULL)
		return;

	vector<Node *> nvec;
	nvec.push_back(root);
	snode.push(root);
	snode.push(NULL);
	int start = 0;
	int end = 1;
	int pos = end;

	while (start != end)
	{
		Node *node = nvec[start];
		cout << node->item << " ";
		start++;

		if (node->right)
		{
			nvec.push_back(node->right);
			snode.push(node->right);
			end++;
		}
		if (node->left)
		{
			nvec.push_back(node->left);
			snode.push(node->left);
			end++;
		}

		if (start == pos)
		{
			pos = end;
			snode.push(NULL);
			cout << endl;
		}
	}
}

利用queue来处理层序遍历

void printLevel(Node *root)
{
	if (root == NULL)
		return;

	queue<Node *> nqueue;
	nqueue.push(root);

	int start = 0;
	int last = 1;
	int tempLast = last;
	while (!nqueue.empty())
	{
		Node *top = nqueue.front();
		nqueue.pop();
		cout << top->data << " ";

		if (top->left)
		{
			nqueue.push(top->left);
			last++;
		}
		if (top->right)
		{
			nqueue.push(top->right);
			last++;
		}

		start++;
		if (start == tempLast)
		{
			tempLast = last;
			cout << endl;
		}
	}
}

关于层序遍历，可以出现一个变种题目，那就是NULL指针打印成空格

void printLevel(Node *root)
{
	if (root == NULL)
		return;

	queue<Node *> qnode;
	qnode.push(root);
	int count = 1;
	int start = 0;
	int tempCount = 0;

	while (!qnode.empty())
	{
		tempCount = 0;
		while (start != count)
		{
			Node *top = qnode.front();
			
			if (top != NULL)
			{
				if (top->left)
				{
					qnode.push(top->left);
					tempCount++;
				}
				else if (top->left == NULL && top->right != NULL)
				{
					qnode.push(NULL);
					tempCount++;
				}
				if (top->right)
				{
					qnode.push(top->right);
					tempCount++;
				}
				else if (top->left != NULL && top->right == NULL)
				{
					qnode.push(NULL);
					tempCount++;
				}
			}

			start++;
			if (top != NULL)
				cout << top->data;
			else
				cout << " ";
			qnode.pop(); 
		}
		count = tempCount;
		start = 0;
		cout << endl;
	}
}