[LeetCode]107. Binary Tree Level Order Traversal II--二叉树层序遍历2

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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]

---

分析：在[LeetCode]107. Binary Tree Level Order Traversal 的基础上，reverse一下最后的结果即可。这里只贴了迭代版本的代码。

/** * Definition for a binary tree node. * 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 == NULL)            return result;        queue<TreeNode*> q;        q.push(root);                   // root不为空，压入队列        int toBePrint = 1;              // 记录本层还需要打印的节点个数        int nextLevel = 0;              // 记录下一层的节点个数，每次入队列加1        vector<int> oneLayer;           // 单层节点        while(!q.empty()){            TreeNode* pCur = q.front();            q.pop();            oneLayer.push_back(pCur->val);            if(pCur->left){             // 左节点不为空，入队列                q.push(pCur->left);                ++nextLevel;            }            if(pCur->right){            // 右结点不为空，入队列                q.push(pCur->right);                ++nextLevel;            }            if(--toBePrint == 0){       // toBePrint 为0，则将oneLayer压入result                toBePrint = nextLevel;                nextLevel = 0;                result.push_back(oneLayer);                oneLayer.clear();       // oneLayer清空            }        }        reverse(result.begin(), result.end());  //反转一下        return result;    }};