Reverse Level Order

• Time : $O(n)$
• Space : $O(n)$

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
void dfs(TreeNode* node, vector<vector<int>>& seq, int dep) {
    if(!node) return;
    if(seq.size() == dep) seq.emplace_back();
    seq[dep].push_back(node->val);
    dfs(node->left, seq, dep + 1);
    dfs(node->right, seq, dep + 1);
}
vector<int> Solution::solve(TreeNode* A) {
    vector<vector<int>> seq;
    dfs(A,seq,0);
    vector<int> res;
    for(int i = seq.size() - 1; i >= 0; i--) {
        for(int j = 0; j < seq[i].size(); j++) {
            res.push_back(seq[i][j]);
        }
    }
    return res;
}