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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] ]
/**
* Definition for a binary tree node.
* 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>> res=new ArrayList<List<Integer>>();
ArrayList<TreeNode> queue=new ArrayList<TreeNode>();
if (root == null){
return res;
}
queue.add(root);
while(!queue.isEmpty()){
List<Integer> values=new ArrayList<Integer>();
ArrayList<TreeNode> subQueue=new ArrayList<TreeNode>();
for(int i=0; i<queue.size(); i++){
values.add(queue.get(i).val);
if(queue.get(i).left != null){
subQueue.add(queue.get(i).left);
}
if(queue.get(i).right != null){
subQueue.add(queue.get(i).right);
}
}
res.add(values);
queue=subQueue;
}
return res;
}
}
LeetCode-Binary Tree Level Order Traversal
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原文地址：http://www.cnblogs.com/incrediblechangshuo/p/5476409.html
