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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.
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 public class Solution {11 public List<List<Integer>> levelOrder(TreeNode root) {12 List<List<Integer>> ret = new ArrayList<List<Integer>>();13 if (root == null) {14 return ret;15 }16 17 Queue<TreeNode> q = new LinkedList<TreeNode>();18 q.offer(root);19 20 while (!q.isEmpty()) {21 int size = q.size();22 List<Integer> list = new ArrayList<Integer>();23 for (int i = 0; i < size; i++) {24 TreeNode cur = q.poll();25 list.add(cur.val);26 27 if (cur.left != null) {28 q.offer(cur.left);29 }30 31 if (cur.right != null) {32 q.offer(cur.right);33 }34 }35 36 ret.add(list);37 }38 39 return ret;40 }41 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/LevelOrder.java
LeetCode: Binary Tree Level Order Traversal 解题报告
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