LeetCode-102. 二叉树的层次遍历

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]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

/*

        3
       / \
      9  20
     / \
    15  7

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

*/


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) {
        if(root==NULL)
            return vec_res;

        tree_queue.push(root);

        while(!tree_queue.empty()){

            /* 每次存储的是一层的数据 */
            int size = tree_queue.size();
            vec_temp.clear();

            while(size--){
                TreeNode* front = tree_queue.front();
                tree_queue.pop();

                if(front->left){
                    tree_queue.push(front->left);
                }
                if(front->right){
                    tree_queue.push(front->right);
                }
                vec_temp.push_back(front->val);
            }
            vec_res.push_back(vec_temp);
        }
        return vec_res;
    }
private:
    vector<int> vec_temp;
    vector<vector<int>> vec_res;
    queue<TreeNode*> tree_queue;
    vector<vector<int>> NodeSaver;
};



int main(){
    //vector<int>srcTree = {3,9,20,NULL,NULL,15,7};

    TreeNode *root = new TreeNode(3);
    TreeNode *l1 = new TreeNode(9);
    TreeNode *r1 = new TreeNode(20);
    TreeNode *ll2 = new TreeNode(15);
    TreeNode *lr2 = new TreeNode(7);

    root->left = l1;
    root->right = r1;
    l1->left = ll2;
    l1->right = lr2;

    Solution *ps = new Solution();
    vector<vector<int>> res = ps->levelOrder(root);
    for(int i=0;i<res.size();i++){
        for(int j=0;j<res[i].size();j++){
            printf("%d ",res[i][j]);
        }
        printf("\n");
    }

    return 0;
}