[LintCode] Binary Tree Level Order Traversal(二叉树的层次遍历)

描述

给出一棵二叉树,返回其节点值的层次遍历(逐层从左往右访问)

样例

给一棵二叉树 {3,9,20,#,#,15,7} :

  3
/ \
9 20
/ \
15 7

返回他的分层遍历结果:

[
[3],
[9,20],
[15,7]
]

挑战

挑战1:只使用一个队列去实现它

挑战2:用BFS算法来做

package com.ossez.lang.tutorial.tests.lintcode;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue; import org.junit.Test;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory; import com.ossez.lang.tutorial.models.TreeNode; /**
* <p>
* 69
* <ul>
* <li>@see <a href=
* "https://www.cwiki.us/display/ITCLASSIFICATION/Binary+Tree+Level+Order+Traversal">https://www.cwiki.us/display/ITCLASSIFICATION/Binary+Tree+Level+Order+Traversal</a>
* <li>@see<a href=
* "https://www.lintcode.com/problem/binary-tree-level-order-traversal">https://www.lintcode.com/problem/binary-tree-level-order-traversal</a>
* </ul>
* </p>
*
* @author YuCheng
*
*/
public class LintCode0069LevelOrderTest { private final static Logger logger = LoggerFactory.getLogger(LintCode0069LevelOrderTest.class); /**
*
*/
@Test
public void testMain() {
logger.debug("BEGIN");
String data = "{3,9,20,#,#,15,7}"; TreeNode tn = deserialize(data);
System.out.println(levelOrder(tn)); } /**
* Deserialize from array to tree
*
* @param data
* @return
*/
private TreeNode deserialize(String data) {
// NULL CHECK
if (data.equals("{}")) {
return null;
} ArrayList<TreeNode> treeList = new ArrayList<TreeNode>(); data = data.replace("{", "");
data = data.replace("}", "");
String[] vals = data.split(","); // INSERT ROOT
TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
treeList.add(root); int index = 0;
boolean isLeftChild = true;
for (int i = 1; i < vals.length; i++) {
if (!vals[i].equals("#")) {
TreeNode node = new TreeNode(Integer.parseInt(vals[i]));
if (isLeftChild) {
treeList.get(index).left = node;
} else {
treeList.get(index).right = node;
}
treeList.add(node);
} // LEVEL
if (!isLeftChild) {
index++;
} // MOVE TO RIGHT OR NEXT LEVEL
isLeftChild = !isLeftChild;
} return root; } private List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> rs = new ArrayList<List<Integer>>(); // NULL CHECK
if (root == null) {
return rs;
} queue.offer(root); while (!queue.isEmpty()) {
int length = queue.size();
List<Integer> list = new ArrayList<Integer>(); for (int i = 0; i < length; i++) {
TreeNode curTN = queue.poll();
list.add(curTN.val);
if (curTN.left != null) {
queue.offer(curTN.left);
}
if (curTN.right != null) {
queue.offer(curTN.right);
}
} rs.add(list);
} return rs;
}
}

点评

这个程序可以使用队列的广度优先算法来进行遍历。

需要注意的是,因为在输出结果的时候需要按照层级来进行输出,那么需要考虑的一个算法就是二叉树的层级遍历算法。

这个算法要求在遍历的时候记录树的层级。

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