Leetcode：102. Binary Tree Level Order Traversal（JAVA）

【问题描述】

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,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
]

【思路】

递归实现

维护一个Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();

层次扫描节点

/**
* 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>> result = new LinkedList<List<Integer>>();
if (root == null) {
return result;
}

//创建一个队列，用于存放所有节点
Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
currentLevel.add(root);
while(!currentLevel.isEmpty()){
//创建List记录当前层节点值
List<Integer> currentList = new LinkedList<Integer>();
int size = currentLevel.size();
for (int i = 0; i < size; i++) {
TreeNode currentNode = currentLevel.poll();
currentList.add(currentNode.val);

if (currentNode.left != null) {
currentLevel.add(currentNode.left);
}
if (currentNode.right != null) {
currentLevel.add(currentNode.right);
}
}
result.add(currentList);
}
return result;
}
}