问题:给定一个二叉树和任意一个节点,求该该节点的下一层的第一个节点,该二叉树为孩子表示法
解法二:
分析:可以考虑先求给定节点的那一层的节点,然后根据该层节点找到下层第一个节点即可,这里我们考虑使用两个队列来实现,先从队列A中移除一个元素,然后将A元素的左右孩子加入到B队列,前提是A的左右孩子不为空,直到A队列中的所有元素都已经移出;然后在对B队列执行类似的操作,直到B队列中的所有元素都已经移出。不断循环这个过程,直到找到给定的元素为止。要注意判空和循环终止条件。
Java实现:
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.LinkedBlockingQueue;
//--------------------- Change Logs----------------------
// <p>@author weiwei.han Initial Created at 2015-9-27<p>
//-------------------------------------------------------
public class FindTreeNode2 {
private static Node root;
public static Node findFirst(Node root, Node e) {
if(root == null || e == null) {
return null;
}
BlockingQueue<Node> queue1 = new LinkedBlockingQueue<>();
BlockingQueue<Node> queue2 = new LinkedBlockingQueue<>();
queue1.offer(root);
boolean flag = true;
while(flag) {
flag = transform(queue1, queue2, e);
if(queue1.isEmpty() && queue2.isEmpty()) {
return null;
}
if(flag) {
flag = transform(queue2, queue1, e);
if(queue1.isEmpty() && queue2.isEmpty()) {
return null;
}
}
}
BlockingQueue<Node> result = queue1.isEmpty() ? queue2 : queue1;
return result.poll();
}
private static boolean transform(BlockingQueue<Node> queue1, BlockingQueue<Node> queue2, Node e) {
boolean flag = true;
while(!queue1.isEmpty()) {
Node data = queue1.poll();
if(data == e) {
flag = false ;
}
if(data.left != null) {
queue2.offer(data.left);
}
if(data.right != null) {
queue2.offer(data.right);
}
}
return flag;
}
private static Node createTree() {
Node node9 = new Node(9, null, null);
Node node8 = new Node(8, null, node9);
Node node6 = new Node(6, node8, null);
Node node3 = new Node(3, null, node6);
Node node7 = new Node(7, null, null);
Node node5 = new Node(5, null, node7);
Node node4 = new Node(4, null, null);
Node node2 = new Node(2, node4, node5);
Node node1 = new Node(1, node2, node3);
root = node1;
return node9;
//return new Node(0, null, null);
//return null;
}
static class Node{
private Object value;
private Node left = null;
private Node right = null;
public Node (Object value, Node left, Node right) {
this.value = value;
this.left = left;
this.right = right;
}
}
public static void main(String[] args) {
Node e = FindTreeNode2.createTree();
Node result = FindTreeNode2.findFirst(root, e);
if(result == null) {
System.out.println("can not find it!!!");
}
else {
System.out.println(result+"--->"+result.value);
}
}
}