LeetCode OJ 之 Binary Tree Level Order Traversal II （二叉树的层次遍历-二）

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题目：

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：

参见二叉树的层次遍历一：http://blog.csdn.net/u012243115/article/details/41091545

只是最后把结果前后反一下即可。

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> > result;        if(root == nullptr)             return result;        queue<TreeNode*> current, next;//定义两个队列，current存当前行的结点，next存储下一行的结点        vector<int> level;         current.push(root);//把根结点入队        while (!current.empty())         {            while (!current.empty()) //结束条件是当前队列为空，下一队列非空            {                TreeNode* node = current.front();                current.pop();                level.push_back(node->val);                if (node->left != nullptr) next.push(node->left);//把当前行结点的子结点都存入next队列                if (node->right != nullptr) next.push(node->right);            }            result.push_back(level);//把当前行的vector push进result            level.clear();//清空当前vector里的数据            swap(next, current);//交换两个队列后，下一行的结点存在current里，next则为空        }        reverse(result.begin(),result.end());//相比之前的层次遍历，只多了这一行        return result;    }};

递归版：

/** * 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;    }    };

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