题解 | 求二叉树的层序遍历

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <queue>
#include <vector>

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector<vector<>>
     */
    vector<vector<int>> result;
    queue<TreeNode*> q;

    void LevelOrderDFS(TreeNode* root, int level){
        if (!root) return;

        if (level == result.size()) {  //动态拓展
            result.push_back(vector<int> {});
        }

        result[level].push_back(root->val);  

        LevelOrderDFS(root->left, level + 1);
        LevelOrderDFS(root->right, level + 1);

    }

    void LevelOrderBFS(TreeNode* root){
        if (!root) return;

        q.push(root);

        while (!q.empty()) {
            vector<int> row;
            int length = q.size();
            for(int i = 0; i < length; ++i){ 
                row.push_back(q.front()->val);
                if (q.front()->left) {q.push(q.front()->left);}
                if (q.front()->right) q.push(q.front()->right);
                q.pop();
            }

            result.push_back(row);
        }
    }

    vector<vector<int> > levelOrder(TreeNode* root) {
        //LevelOrderDFS(root, 0);  //深度优先搜索的方法

        LevelOrderBFS(root);  //广度优先搜索的方法

        return result;
    }
};