[LeetCode] 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.

 

 

在 [LeetCode] Binary Tree Level Order Traversal的基础上，将result reverse 一下就OK

class Solution {
    public:
        vector<vector<int> > levelOrderBottom(TreeNode *root)
        {
            queue<TreeNode*> q1;
            queue<TreeNode*> q2;
            vector<vector<int> > res;

            int depth = 0;
            if(root != NULL)
            {
                q1.push(root);
                res.push_back(vector<int> ());
            }

            while(!q1.empty())
            {
                TreeNode * p = q1.front();
                q1.pop();

                res[depth].push_back(p->val);

                if(p->left)
                    q2.push(p->left);
                if(p->right)
                    q2.push(p->right);
                if(q1.empty() && !q2.empty())
                {
                    swap(q1, q2);
                    depth ++;
                    res.push_back(vector<int> ());
                }
            }   
            reverse(res.begin(), res.end());
            return res;
        }
};