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 = 0; 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;
    }
};