LeetCode: Binary Tree Level Order Traversal 解题报告

Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

SOLUTION 1:

使用队列来解决，很直观。注意在每次换层的时候，新建一个List.

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<List<Integer>> levelOrder(TreeNode root) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         if (root == null) {
             return ret;
         }

         Queue<TreeNode> q = new LinkedList<TreeNode>();
         q.offer(root);

         while (!q.isEmpty()) {
             int size = q.size();
             List<Integer> list = new ArrayList<Integer>();
             for (int i = 0; i < size; i++) {
                 TreeNode cur = q.poll();
                 list.add(cur.val);

                 if (cur.left != null) {
                     q.offer(cur.left);
                 }

                 if (cur.right != null) {
                     q.offer(cur.right);
                 }
             }

             ret.add(list);
         }

         return ret;
     }
 }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/LevelOrder.java