Binary Tree Level Order Traversal ---- LeetCode
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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]]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
解题思路: 使用两个队列来完成层次遍历,其中一个对列存放当前遍历的层次节点,另一存放下一层节点即当前遍历的节点的孩子节点,当前对列为空则调换两个队列继续遍历。
/**
* 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>> rs=new ArrayList<List<Integer>>();
Queue<TreeNode> queue1=new LinkedList<TreeNode>();
Queue<TreeNode> queue2=new LinkedList<TreeNode>();
if(root==null){
return rs;
}
queue1.add(root);
while(!queue1.isEmpty()){
List<Integer> list=new ArrayList<Integer>();
while(!queue1.isEmpty()){
TreeNode node=queue1.poll();
if(node.left!=null){
queue2.offer(node.left);
}
if(node.right!=null){
queue2.offer(node.right);
}
list.add(node.val);
}
rs.add(list);
queue1.addAll(queue2);
queue2.clear();
}
return rs;
}
}
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