Binary Tree Level Order Traversal II 从下向上按层遍历二叉树

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]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > vec;        queue<TreeNode *> cur;        if(root==NULL)            return vec;        cur.push(root);                vector<int> vv;         queue<TreeNode *> next;                while(!cur.empty())        {                TreeNode *t = cur.front();            cur.pop();            vv.push_back(t->val);            if(t->left)                next.push(t->left);            if(t->right)                next.push(t->right);                            if(cur.empty())            {                 vec.insert(vec.begin(),vv);//vv从头插入                // cur=next;                //next.clear();                swap(cur,next);                 vv.clear();            }        }                return vec;        }};