[leetcode] 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]]

思路：宽度优先搜索，使用队列，与Binary Tree Level Order Traversal类似，最后将结果反转即可

代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<vector<int> > levelOrderBottom(TreeNode *root) {        vector<vector<int> > res;        if(root==NULL) return res;        vector<int> temp;        queue<TreeNode*> wfs;        wfs.push(root);        int level=0;        int count=1;        TreeNode *tmp;        while(!wfs.empty()){            temp.clear();            level=0;            for(int i=0;i<count;i++){                tmp=wfs.front();                wfs.pop();                temp.push_back(tmp->val);                if(tmp->left!=NULL){                    wfs.push(tmp->left);                    level++;                }                if(tmp->right!=NULL){                    wfs.push(tmp->right);                    level++;                }            }            count=level;            res.push_back(temp);        }        reverse(res.begin(),res.end());        return res;    }};