leetcode107 二叉树的层次遍历2

只需要将正常层次遍历的结果翻转即可

遇到的问题：在 level 函数中，res 是按值传递的，这意味着在递归过程中对 res 的修改不会影响外部的 res 变量。因此，最终返回的 res 始终是空的，即使你在 level 函数里添加了元素。

/**
 * 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 {
private:
    void level(TreeNode* cur, int num, vector<vector<int>>& res){
        if(cur == nullptr) return;
        vector<int> curlevel;
        if(num > res.size()) res.push_back(vector<int> ());
        res[num-1].push_back(cur->val);
        if(cur->left) level(cur->left, num + 1, res);
        if(cur->right) level(cur->right, num + 1, res);
    }
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        int num = 1;
        level(root, num, res);
        reverse(res.begin(), res.end());
        return res;
    }
};

小问题：if(!q.empty()) 这个条件会导致只处理一层，而不会处理所有层。正确的做法应该是用 while(!q.empty()) 循环处理每一层。

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        int levelnum;
        queue<TreeNode*> q;
        if(root == nullptr) return res;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            vector<int> everylevel;
            
            for(int i = 0; i < size; i++){
                TreeNode* cur = q.front();
                q.pop();
                everylevel.push_back(cur->val);
                if(cur->left) q.push(cur->left);
                if(cur->right) q.push(cur->right);
            }

            res.push_back(everylevel);
        }
        reverse(res.begin(), res.end());
        return res;
    }
};