题目

binary-tree-level-order-traversal-ii

2. 算法

* 非递归

* 递归

3. 代码

* 非递归

// Iterative
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > res;
        if (root == NULL) return res;

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

* 递归

// Recurive
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);
    }
};