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] ]
方法1:深度优先(递归算法)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
levelHelper(res, root, 0);
return res;
}
public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
if (root == null) return;
if (height >= res.size()) {
res.add(new LinkedList<Integer>());
}
res.get(height).add(root.val);
levelHelper(res, root.left, height+1);
levelHelper(res, root.right, height+1);
}
}
时间复杂度:O(n)
空间复杂度:O(n)
方法2:一种非递归的方式,利用栈的形式
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
Queue<TreeNode> queue=new LinkedList<>();
queue.add(root);
while(!queue.isEmpty()){
List<Integer> list=new ArrayList();
int size=queue.size();
for(int i=0;i<size;i++){
TreeNode p=queue.poll();
list.add(p.val);
if(p.left!=null){
queue.add(p.left);
}
if(p.right!=null){
queue.add(p.right);
}
}
res.add(list);
}
return res;
}
}
时间复杂度:O(n)
空间复杂度:O(n)