Binary Tree Level Order Traversal II

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],]

 

 

/** * 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> > out;    void levelTraval(vector<TreeNode* >& level)    {        if(level.size() > 0)        {            vector<TreeNode*> nextLevel;            for(int i = 0; i < level.size(); i++)            {                if(level[i]->left)                    nextLevel.push_back(level[i]->left);                if(level[i]->right)                    nextLevel.push_back(level[i]->right);            }            levelTraval(nextLevel);            vector<int> val;            for(int i = 0; i < level.size(); i++)                val.push_back(level[i]->val);            out.push_back(val);        }    }    vector<vector<int> > levelOrderBottom(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        out.clear();        vector<TreeNode*> v;        if(root != NULL)        {            v.push_back(root);            levelTraval(v);        }        return out;    }};