题解 | #从下到上打印二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > levelOrderBottom(TreeNode* root) {
        // write code here
        vector<vector<int>> res;
        vector<int> ans;
        if(!root)
            return res;
        ans.push_back(root->val);
        res.push_back(ans);
        queue<TreeNode*> cel;
        if(root->left)
            cel.push(root->left);
        if(root->right)
            cel.push(root->right);
        int count = cel.size();
        ans.clear();
        while(cel.size()) {
            auto r = cel.front();
            cel.pop();
            count--;
            ans.push_back(r->val);
            if(r->left)
                cel.push(r->left);
            if(r->right)
                cel.push(r->right);
            
            if(count == 0) {
                count = cel.size();
                res.push_back(ans);
                ans.clear();
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};