Cinte's Leetcode Record
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102. Binary Tree Level Order Traversal

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Symbols count in article: 1.7k Reading time ≈ 2 mins.

102. Binary Tree Level Order Traversal

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if(!root) return {};
        vector<vector<int>> ret;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            vector<int> tmp;
            for(int i = 0, n = q.size(); i < n; ++i)
            {
                auto p = q.front();
                q.pop();
                tmp.push_back(p->val);
                if(p->left)
                    q.push(p->left);
                if(p->right)
                    q.push(p->right);
            }
            ret.push_back(tmp);
        }
        return ret;
    }
};
image.png

还有使用递归，传入层数，对每一层的vector容器加入

review 

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if(!root)
            return {};
        vector<vector<int>> ret;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            ret.emplace_back();
            for(int i = 0, sz = q.size(); i < sz; ++i)
            {
                auto p = q.front();
                q.pop();
                ret.back().push_back(p->val);
                if(p->left)
                    q.push(p->left);
                if(p->right)
                    q.push(p->right);
            }
        }
        return ret;
    }
};