文章作者：Tyan
博客：noahsnail.com ?|? CSDN ?|? 簡書
1. 問題描述
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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
2. 求解
這個題就是一個樹的層次遍歷問題颠焦，需要用到新的數(shù)據(jù)結(jié)構(gòu)隊列憾股，把每一層的結(jié)點的子結(jié)點放入到隊列中，依次遍歷黎泣。
/**
* 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>> list = new ArrayList<List<Integer>>();
if(root == null) {
return list;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
Queue<TreeNode> result = new LinkedList<TreeNode>();
List<Integer> level = new ArrayList<Integer>();
while(!queue.isEmpty()) {
TreeNode temp = queue.poll();
level.add(temp.val);
if(temp.left != null) {
result.add(temp.left);
}
if(temp.right != null) {
result.add(temp.right);
}
if(queue.isEmpty()) {
queue = result;
result = new LinkedList<TreeNode>();
list.add(level);
level = new ArrayList<Integer>();
}
}
return list;
}
}
