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M
方法1. 最普通,Non-recursive: BFS, queue, 用个queue.size()来end for loop:换行。
或者用两个queue. 当常规queue empty，把backup queue贴上去。
方法2. Recursive with dfs:
每个level都应该有个ArrayList. 那么用一个int level来查看：是否每一层都有了相应的ArrayList。
如果没有，就加上一层。
之后每次都通过DFS在相应的level上面加数字。
```
/*
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
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]
]
Challenge
Challenge 1: Using only 1 queue to implement it.
Challenge 2: Use DFS algorithm to do it.
Tags Expand
Queue Binary Tree Breadth First Search Binary Tree Traversal Uber LinkedIn Facebook
*/
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
/*
Thoughts:
1. Non-recursive
Use queue to withhold the parent.
Poll one parent, add this parent’s value to arrayList
Add the children into Arraylist
jump to next level
2. Recursive
use a integer to track levels.
If at a new level, then create a new ArrayList.
At each node, add the node to corresponding level-ArrayList
*/
//Non-recurive Iterative way:
//Even with while + for nested loop, it's just O(n)
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (root == null) {
return result;
}
//Use a queue to list elements: each row
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
ArrayList<Integer> list = new ArrayList<Integer>();
int size = queue.size();//Limit the size, since the queue is increasing
for (int i = 0; i < size; i++) {
TreeNode levelNode = queue.poll();
list.add(levelNode.val);//Add all the values from this current level
if (levelNode.left != null) {
queue.offer(levelNode.left);
}
if (levelNode.right != null) {
queue.offer(levelNode.right);
}
}
result.add(list);
}//while
return result;
}
}
//Another Iterative way: using 2 Queues
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> rst = new ArrayList<List<Integer>>();
if (root == null) {
return rst;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
Queue<TreeNode> backQueue = new LinkedList<TreeNode>();
queue.offer(root);
ArrayList<Integer> list = new ArrayList<Integer>();
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if (node.left != null) {
backQueue.offer(node.left);
}
if (node.right != null) {
backQueue.offer(node.right);
}
list.add(node.val);
if (queue.isEmpty()) {
rst.add(new ArrayList<Integer>(list));
list = new ArrayList<Integer>();
queue = backQueue;
backQueue = new LinkedList<TreeNode>();
}
}
return rst;
}
}
//Recursive:
//Recursive with dfs: use a level to track. Add curr into corresponding level; each level > rst.size(), add a new [].
//Note: rst is a ArrayList<ArrayList<Integer>>, where each level is a arraylist; that is why we can add [] into rst to represent a level.
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (root == null) {
return result;
}
dfs(root, 0, result);
return result;
}
public void dfs(TreeNode root, int level, ArrayList<ArrayList<Integer>> rst) {
if (root == null) {
return;
}
if (level >= rst.size()) {
rst.add(new ArrayList<Integer>());
}
rst.get(level).add(root.val);
dfs(root.left, level + 1, rst);
dfs(root.right, level + 1, rst);
}
}
```