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<vector<int>> levelOrderBottom(TreeNode* root)    {        //同二叉树的层次遍历        vector<vector<int>>result;        if (!root)            return result;        queue<TreeNode*> vec;        vec.push(root);        while (vec.size() > 0) {            queue<TreeNode*> tmp_vec;            vector<int> 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;    }};