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