lintcode：二叉树的层次遍历

地址：

http://lintcode.com/zh-cn/problem/binary-tree-level-order-traversal/

借助队列来完成

class Solution {
public:
    /*
     * @param root: A Tree
     * @return: Level order a list of lists of integer
     */
    vector<vector<int>> levelOrder(TreeNode * root) {
        // write your code here
        vector<vector<int>> res;
        if(root==NULL)
            return res;
        queue<TreeNode*> queue;
        queue.push(root);
        
        while(!queue.empty()){
            vector<int> cur;
            int len = queue.size();  
            
            while(len--){
                 TreeNode *tmp=queue.front();  
                 cur.push_back(tmp->val);
                 
                 queue.pop();
                 
                 if(tmp->left)
                    queue.push(tmp->left);
                
                 if(tmp->right)
                    queue.push(tmp->right);
                 
            }
            res.push_back(cur);
        }
        return res;
    }
};

http://lintcode.com/zh-cn/problem/binary-tree-level-order-traversal-ii/

给出一棵二叉树，返回其节点值从底向上的层次序遍历（按从叶节点所在层到根节点所在的层遍历，然后逐层从左往右遍历）

这个题目是从底向上，其实类似上题，主要是因为用了vector方便很多:

class Solution {
public:
    /*
     * @param root: A tree
     * @return: buttom-up level order a list of lists of integer
     */
    vector<vector<int>> levelOrderBottom(TreeNode * root) {
        // write your code here
        queue<TreeNode*> queue;
        vector<vector<int>> res;
        int len;
        if(root==NULL)
            return res;
        queue.push(root);

        while(!queue.empty()){
            len = queue.size();
            vector<int> cur;
            while(len--){
                TreeNode* temp = queue.front();
                cur.push_back(temp->val);
                queue.pop();
                
                if(temp->left){
                    queue.push(temp->left);
                }
                
                if(temp->right){
                    queue.push(temp->right);
                }
            }
            if(!cur.empty())
            res.insert(res.begin(),cur);
        }
        return res;
    }
};