package com.lever;
import java.util.LinkedList;
import java.util.Queue;
/**
* 二叉树遍历
* @author lckxxy
*
*/
public class Node {
public int value;
public Node left;
public Node right;
public Node(int v) {
this.value = v;
this.left = null;
this.right = null;
}
public static void main(String[] args) {
StringBuffer str = new StringBuffer();
str.append("hello world");
String test = str.toString();
System.out.println(test.replace(" ", "%20"));
Node n1 = new Node(1);
Node n2 = new Node(2);
Node n3 = new Node(3);
Node n4 = new Node(4);
Node n5 = new Node(5);
Node n6 = new Node(6);
Node n7 = new Node(7);
Node n8 = new Node(8);
Node n9 = new Node(9);
Node n10 = new Node(10);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n3.left = n6;
n3.right = n7;
n4.left = n8;
n4.right = n9;
n5.left = n10;
levelTravel(n1);
}
/**
*
* @param root
* 树根节点
* 层序遍历二叉树，用队列实现，先将根节点入队列，只要队列不为空，然后出队列，并访问，接着讲访问节点的左右子树依次入队列
*/
public static void levelTravel(Node root) {
if (root == null)
return;
// 当前行最后一个元素
Node last = root;
// 下一行最后一个元素
Node nlast = root.right == null ? root.left : root.right;
Queue q = new LinkedList();
q.add(root);
while (!q.isEmpty()) {
Node temp = (Node) q.poll();
System.out.print(temp.value + " ");
if (temp.left != null) {
q.add(temp.left);
// 如果左儿子不为空，把nlast置为左儿子，这样保证nlast永远是当前行最新一个
nlast = temp.left;
}
if (temp.right != null) {
q.add(temp.right);
// 如果右儿子不为空，把nlast置为右儿子
nlast = temp.right;
}
// 如果当前node跟last相同，说明该换行了，输出换行符，并将nlast的赋给last
if (temp.equals(last)) {
System.out.println();
last = nlast;
}
}
}
}