给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。
例如:
给定二叉树: [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
返回其层次遍历结果:
[ [3], [9,20], [15,7] ]
方法:首先把每层的链表放入queue中进行遍历,并将每层的链表内容存入vector中,最后
将每层的vector存入大的vector内
/**
* 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>> all;
vector<int > level;
queue<TreeNode*> que;
TreeNode* tree;
if (root == NULL) return all;
que.push(root);
while (!que.empty())
{
level.clear();
int size = que.size();
for (int i = 0; i < size; i++){
tree = que.front();
que.pop();
level.push_back(tree->val);
if (tree->left)
que.push(tree->left);
if (tree->right)
que.push(tree->right);
}
all.push_back(level);
}
return all;
}
};