102. Binary Tree Level Order Traversal

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]
]

思路

典型的BFS的题目，维护一个队列，将每层的节点都依次入队，入队完后记录每层的数量。

c++11

方法一：迭代

/**
 * 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) {
            queue<TreeNode*> q;
            q.emplace(root);
            while(!q.empty()) {
                vector<int> v;
                int cur_layer_size = q.size();
                for (int i = 0; i < cur_layer_size; ++i) {
                    TreeNode* temp = q.front();
                    q.pop();
                    v.emplace_back(temp->val);
                    if(temp->left) q.emplace(temp->left);
                    if(temp->right) q.emplace(temp->right);
                }
                res.emplace_back(v);
            }
        }
        return res;
    }
};

方法二：递归

/**
 * 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;
        BFS(root, 0, res);
        return res;
    }

    void BFS(TreeNode* root, int level, vector<vector<int>> &res) {
        if(!root) return;
        if(res.size() == level) res.push_back({});
        res[level].push_back(root->val);

        if(root->left) BFS(root->left, level+1, res);
        if(root->right) BFS(root->right, level+1, res);
    }
};