【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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

code: 1A 36ms

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<vector<int> > res;
        if(root == NULL)
            return res;
        queue<TreeNode *> qu;
        qu.push(root);
        qu.push(NULL);
        vector<int> onelevel;
        while(true)
        {
            TreeNode *cur = qu.front();
            qu.pop();
            if(cur == NULL)
            {
                res.push_back(onelevel);
                onelevel.clear();
                if(qu.empty())
                    break;
                qu.push(NULL);
            }
            else
            {
                onelevel.push_back(cur->val);
                if(cur->left)
                    qu.push(cur->left);
                if(cur->right)
                    qu.push(cur->right);
            }
            
        }
        reverse(res.begin(),res.end());
        return res;
    }
};