Binary Tree Level Order Traversal II

题目描述：

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]
]

思路：先层序遍历二叉树，再将得到的结果翻转。

代码：

vector<vector<int> > levelOrderBottom(TreeNode *root) {
    vector<vector<int> > result;
    vector<int> sequence;
    TreeNode * flag = new TreeNode(0);
    queue<TreeNode *> treenode_queue;
    if(root == NULL)
        return result;
    treenode_queue.push(root);
    treenode_queue.push(flag);
    while(treenode_queue.size() >= 1)
    {
        TreeNode * node = treenode_queue.front();
        treenode_queue.pop();
        if(node == flag)
        {
            result.push_back(sequence);
            sequence.clear();
            if(treenode_queue.empty())
                break;
            treenode_queue.push(flag);
        }
        else
        {
            sequence.push_back(node->val);
            if(node->left != NULL)
                treenode_queue.push(node->left);
            if(node->right != NULL)
                treenode_queue.push(node->right);
        }
    }
    reverse(result.begin(),result.end());
    return result;
    }

Binary Tree Level Order Traversal II

原文：http://blog.csdn.net/yao_wust/article/details/41241985