【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]
]

思路：如果直接从题目要求考虑的话可以会用到更多的空间，比如需要记录每一层的节点个数，需要用一个queue和一个stack来保存数据，但是如果是从【Leetcode】Binary
Tree Level Order Traversal开始考虑，就会简单的多，只要将结果进行下reverse即可。

/**
 * 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) {
        vector<vector<int>> result;

        if(root == NULL)    return result;

        queue<TreeNode *> lq;
        lq.push(root);
        int curlc = 1;
        int nextlc = 0;

        while(!lq.empty())
        {
            vector<int> level;
            while(curlc > 0)
            {
                TreeNode * temp = lq.front();
                lq.pop();
                curlc--;
                level.push_back(temp->val);

                if(temp->left)
                {
                    lq.push(temp->left);
                    nextlc++;
                }
                if(temp->right)
                {
                    lq.push(temp->right);
                    nextlc++;
                }
            }

            curlc = nextlc;
            nextlc = 0;
            result.push_back(level);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};

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