数据结构——树的水平遍历

原文地址：Level Order Tree Traversal

树的水平遍历就是广度优先遍历树。

上树的水平遍历顺序是：1 2 3 4 5。

方法1 （用函数打印指定的层）

算法：这个方法里面有两个基本的函数。一个是打印已知层的所有节点（printGivenLevel），另一个是打印树的水平顺序遍历（printLevelorder）。printLevelorder用printGivenLevel逐个从根部打印所有层的节点。

/*Function to print level order traversal of tree*/printLevelorder(tree)for d = 1 to height(tree)   printGivenLevel(tree, d);/*Function to print all nodes at a given level*/printGivenLevel(tree, level)if tree is NULL then return;if level is 1, then    print(tree->data);else if level greater than 1, then    printGivenLevel(tree->left, level-1);    printGivenLevel(tree->right, level-1);

实现：

// Recursive Java program for level order traversal of Binary Tree/* Class containing left and right child of current    node and key value*/class Node{    int data;    Node left, right;    public Node(int item)    {        data = item;        left = right = null;    }}class BinaryTree{    // Root of the Binary Tree    Node root;    public BinaryTree()    {        root = null;    }    /* function to print level order traversal of tree*/    void printLevelOrder()    {        int h = height(root);        int i;        for (i=1; i<=h; i++)            printGivenLevel(root, i);    }    /* Compute the "height" of a tree -- the number of    nodes along the longest path from the root node    down to the farthest leaf node.*/    int height(Node root)    {        if (root == null)           return 0;        else        {            /* compute  height of each subtree */            int lheight = height(root.left);            int rheight = height(root.right);            /* use the larger one */            if (lheight > rheight)                return(lheight+1);            else return(rheight+1);         }    }    /* Print nodes at the given level */    void printGivenLevel (Node root ,int level)    {        if (root == null)            return;        if (level == 1)            System.out.print(root.data + " ");        else if (level > 1)        {            printGivenLevel(root.left, level-1);            printGivenLevel(root.right, level-1);        }    }    /* Driver program to test above functions */    public static void main(String args[])    {       BinaryTree tree = new BinaryTree();       tree.root= new Node(1);       tree.root.left= new Node(2);       tree.root.right= new Node(3);       tree.root.left.left= new Node(4);       tree.root.left.right= new Node(5);       System.out.println("Level order traversal of binary tree is ");       tree.printLevelOrder();    }}

输出：

Level order traversal of binary tree is - 1 2 3 4 5 

最坏情况下的时间复杂度是：O(n^2) ，对于一个偏树（skewed tree），printGivenLevel()花费的时间是O(n)，n在这里是偏树的节点数目。所以printLevelOrder()的时间复杂度是O(n) + O(n-1) + O(n-2) + .. + O(1)，也就是O(n^2)

方法2：（用队列）

算法：

对于每一个节点，首先遍历这个节点，然后将这个节点的孩子节点推到一个FIFO的队列。

printLevelorder(tree)1) Create an empty queue q2) temp_node = root /*start from root*/3) Loop while temp_node is not NULL    a) print temp_node->data.    b) Enqueue temp_node’s children (first left then right children) to q    c) Dequeue a node from q and assign it’s value to temp_node

实现：

这里对上述算法做了一个简单的实现。队列是用数组实现的，它的最大尺寸是500。我们也可以用链表实现一个队列。

// Iterative Queue based Java program to do level order traversal// of Binary Tree/* importing the inbuilt java classes required for the program */import java.util.Queue;import java.util.LinkedList;/* Class to represent Tree node */class Node {    int data;    Node left, right;    public Node(int item) {        data = item;        left = null;        right = null;    }}/* Class to print Level Order Traversal */class BinaryTree {    Node root;    /* Given a binary tree. Print its nodes in level order     using array for implementing queue  */    void printLevelOrder()     {        Queue<Node> queue = new LinkedList<Node>();        queue.add(root);        while (!queue.isEmpty())         {            /* poll() removes the present head.            For more information on poll() visit             http://www.tutorialspoint.com/java/util/linkedlist_poll.htm */            Node tempNode = queue.poll();            System.out.print(tempNode.data + " ");            /*Enqueue left child */            if (tempNode.left != null) {                queue.add(tempNode.left);            }            /*Enqueue right child */            if (tempNode.right != null) {                queue.add(tempNode.right);            }        }    }    public static void main(String args[])     {        /* creating a binary tree and entering          the nodes */        BinaryTree tree_level = new BinaryTree();        tree_level.root = new Node(1);        tree_level.root.left = new Node(2);        tree_level.root.right = new Node(3);        tree_level.root.left.left = new Node(4);        tree_level.root.left.right = new Node(5);        System.out.println("Level order traversal of binary tree is - ");        tree_level.printLevelOrder();    }}

输出：

Level order traversal of binary tree is - 1 2 3 4 5 

时间复杂度：O(n)，n在这里是二叉树的节点数。