LeetCode107 Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). （Easy）

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

分析：

只需要在层序遍历的基础上（参见Binary Tree Level Order Traversal）将最后结果vector reverse即可

代码：

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 private:
     vector<vector<int>> result;
 public:
     vector<vector<int>> levelOrderBottom(TreeNode* root) {
         if (root == nullptr) {
             return result;
         }
         queue<TreeNode* > que;
         que.push(root);
         while (!que.empty()) {
             int sz = que.size();
             vector<int> temp;
             for (int i = ; i < sz; ++i) {
                 TreeNode* cur = que.front();
                 que.pop();
                 temp.push_back(cur -> val);
                 if (cur -> left != nullptr) {
                     que.push(cur -> left);
                 }
                 if (cur -> right != nullptr) {
                     que.push(cur -> right);
                 }
             }
             result.push_back(temp);
         }
         reverse(result.begin(), result.end());
         return result;
     }
 };