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).

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]
]

分析：

用两个ArrayList,一个保存当前这行所有的Node, 然后另一个数组保存下一行所有的nodes.

 1 /**
 2  * Definition of TreeNode:
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left, right;
 6  *     public TreeNode(int val) {
 7  *         this.val = val;
 8  *         this.left = this.right = null;
 9  *     }
10  * }
11  */
12
13
14 public class Solution {
15     /**
16      * @param root: The root of binary tree.
17      * @return: Level order a list of lists of integer
18      */
19     public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
20         ArrayList<ArrayList<Integer>> allLists = new ArrayList<>();
21         if (root == null) return allLists;
22
23         ArrayList<Integer> nums = new ArrayList<>();
24         ArrayList<TreeNode> list1 = new ArrayList<>();
25         ArrayList<TreeNode> list2 = new ArrayList<>();
26
27         list1.add(root);
28
29         while(list1.size() != 0) {
30             for (TreeNode node : list1) {
31                 nums.add(node.val);
32                 if (node.left != null) list2.add(node.left);
33                 if (node.right != null) list2.add(node.right);
34             }
35             list1.clear();
36             allLists.add(new ArrayList<Integer>(nums));
37             nums.clear();
38             ArrayList<TreeNode> temp = list1;
39             list1 = list2;
40             list2 = temp;
41         }
42         return allLists;
43     }
44 }

转载于:https://www.cnblogs.com/beiyeqingteng/p/5649664.html

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