leetcode第47題(binary-tree-level-order-traversal)
題目:
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] ]
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
思路:
跟前面42題一樣,只是最後加的時候不加在頭部即可。
程式碼:
importjava.util.ArrayList; import java.util.Collections; import java.util.LinkedList; import java.util.Queue; /** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { publicArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if(root==null){ return result; } Queue<TreeNode> que = new LinkedList<TreeNode>(); que.add(root); while(!que.isEmpty()){ int count = que.size(); ArrayList<Integer> list = new ArrayList<Integer>(); for(int i=0;i<count;i++){ TreeNode node = que.poll(); list.add(node.val); if(node.left!=null){ que.add(node.left); } if(node.right != null){ que.add(node.right); } } result.add(list); } return result; } }
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] ]
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".