给你一个二叉树,请你返回其按 层序遍历 得到的节点值。 (即逐层地,从左到右访问所有节点)。
示例:
二叉树:[3,9,20,null,null,15,7]
返回其层次遍历结果:
C++代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>>res;
if(root==NULL)
return res;
queue<TreeNode*>q;
q.push(root);
while(!q.empty())
{
vector<int>r;
int size=q.size();
for(int i=0;i<size;i++)
{
TreeNode* t=q.front();
r.push_back(t->val);
q.pop();
if(t->left)
q.push(t->left);
if(t->right)
q.push(t->right);
}
res.push_back(r);
}
return res;
}
};
JAVA代码如下:
/**
* 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>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> list=new LinkedList<>();
if(root==null){
return list;
}
Queue<TreeNode> que=new LinkedList<>();
que.offer(root);
while (!que.isEmpty()){
List<Integer> r=new ArrayList<>();
int size=que.size();
for (int i = 0; i < size; i++) {
TreeNode t=que.poll();
r.add(t.val);
if(t.left!=null){
que.offer(t.left);
}
if(t.right!=null){
que.offer(t.right);
}
}
list.addFirst(r);
}
return list;
}
}