• [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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  • 原文地址:https://www.cnblogs.com/huyuchengus/p/10130772.html
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