LeetCode OJ 之 Binary Tree Level Order Traversal II （二叉樹的層次周遊-二）

題目：

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：

參見二叉樹的層次周遊一：http://blog.csdn.net/u012243115/article/details/41091545

隻是最後把結果前後反一下即可。

代碼：

/**
 * 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> > result;
        if(root == nullptr) 
            return result;
        queue<TreeNode*> current, next;//定義兩個隊列，current存目前行的結點，next存儲下一行的結點
        vector<int> level; 
        current.push(root);//把根結點入隊
        while (!current.empty()) 
        {
            while (!current.empty()) //結束條件是目前隊列為空，下一隊列非空
            {
                TreeNode* node = current.front();
                current.pop();
                level.push_back(node->val);
                if (node->left != nullptr) next.push(node->left);//把目前行結點的子結點都存入next隊列
                if (node->right != nullptr) next.push(node->right);
            }
            result.push_back(level);//把目前行的vector push進result
            level.clear();//清空目前vector裡的資料
            swap(next, current);//交換兩個隊列後，下一行的結點存在current裡，next則為空
        }
        reverse(result.begin(),result.end());//相比之前的層次周遊，隻多了這一行
        return result;
    }
};
           

遞歸版：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void traverse(TreeNode *root, size_t level, vector<vector<int>> &result) 
    {
        if (!root) 
            return;
        if (level > result.size())
        result.push_back(vector<int>());
        
        result[level-1].push_back(root->val);
        
        traverse(root->left, level+1, result);
        traverse(root->right, level+1, result);

    }
    vector<vector<int> > levelOrderBottom(TreeNode *root) 
    {
        vector<vector<int>> result;
        traverse(root, 1, result);
        std::reverse(result.begin(), result.end()); 
        return result;
    }
    
};