题目

题解

二叉树结构如下：

//Definition for a binary tree node.
    struct TreeNode {
        int val;
        TreeNode *left;
        TreeNode *right;
        TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    };

1. 队列迭代法

比上一题再复杂一点，要区分奇偶行，偶数行从后面加数，奇数行从前面加数，详见103题

  • 时间复杂度$O(n)$
  • 空间复杂度$O(n)$
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(!root) return res;
        queue<TreeNode*> q({root});
        while(!q.empty()){
            vector<int> thisLevel;
            int levelSize = q.size();
            while(levelSize--){
                auto tmp = q.front();q.pop();
                if(res.size()%2==0) thisLevel.push_back(tmp->val);//偶数列
                else thisLevel.insert(thisLevel.begin(), tmp->val);//奇数列
                if(tmp->left) q.push(tmp->left);
                if(tmp->right) q.push(tmp->right);
            }
            res.push_back(thisLevel);
        }
        return res;
    }
};

2. 递归法

详见103题注释

  • 时间复杂度$O(n)$
  • 空间复杂度$O(n)$
class Solution {
    vector<vector<int>> res;
public:
    void dfs(TreeNode* cur, int level){
        if(!cur) return;
        if(level == res.size()) res.push_back({});
        if(level%2==0) res[level].push_back(cur->val);
        else res[level].insert(res[level].begin(),cur->val);
        dfs(cur->left, level+1);
        dfs(cur->right, level+1);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        if(!root) return res;
        dfs(root,0);
        return res;
    }
};