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binary-tree-level-order-traversal-ii

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Binary Tree Level Order Traversal II – LeetCode 107

Problem

Description

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

Answer

Original

Code

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/**
 * 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>> levelOrderBottom(TreeNode* root) {
        if(!root) return vector<vector<int>>();
        queue<TreeNode *> q;
        vector<vector<int>> result;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            vector<int> now;
            for(int i = 0; i != size; ++i){
                TreeNode *tmp = q.front(); q.pop();
                now.push_back(tmp->val);
                if(tmp->left) { q.push(tmp->left);}
                if(tmp->right) {q.push(tmp->right);}
            }
            result.insert(result.begin(),now);   
        }
        return result;
    }
};

思路

因为要求一层一层的输出，于是只好用BFS写，时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$6$ ms，排名$53.42\%$

Better

思路

递归解，妙在每层之间对于res的处理，值得借鉴，时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$6$ ms，排名$53.42\%$

Code

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/**
 * 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>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > res;
        levelorder(root, 0, res);
        return vector<vector<int> > (res.rbegin(), res.rend());
    }
    void levelorder(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) levelorder(root->left, level + 1, res);
        if (root->right) levelorder(root->right, level + 1, res);
    }
};