LeetCode 107: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).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

此题与LeetCode 101  Binary Tree Level Order Traversal，一样都是二叉树的层次遍历，不同的是这一次是从叶子，到根节点。所以只需要在LeetCode101的基础上再做一次翻转即可。代码如下：

class Solution
{
public:
    vector> levelOrderBottom(TreeNode* root)
    {
        //同二叉树的层次遍历
        vector>result;
        if (!root)
            return result;
        queue vec;
        vec.push(root);
        while (vec.size() > 0) {
            queue tmp_vec;
            vector tmp_result;
            while (vec.size() > 0) {
                TreeNode* node = vec.front();
                vec.pop();
                if (node->left)
                    tmp_vec.push(node->left);
                if (node->right)
                    tmp_vec.push(node->right);
                tmp_result.push_back(node->val);
            }

            vec = tmp_vec;
            result.push_back(tmp_result);
        }
        //对层次遍历结果进行翻转
        reverse(result.begin(), result.end());
        return result;
    }
};