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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] ]
解题思路：
前序遍历(BFS)，使用一个Queue存储每层元素即可，JAVA实现如下：
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
if (root == null)
return list;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (queue.size() != 0) {
List<Integer> alist = new ArrayList<Integer>();
for (TreeNode child : queue)
alist.add(child.val);
list.add(new ArrayList<Integer>(alist));
Queue<TreeNode> queue2=queue;
queue=new LinkedList<TreeNode>();
for(TreeNode child:queue2){
if (child.left != null)
queue.add(child.left);
if (child.right != null)
queue.add(child.right);
}
}
return list;
}
Java for LeetCode 102 Binary Tree Level Order Traversal
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原文地址：http://www.cnblogs.com/tonyluis/p/4523689.html
