leetcode Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

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Total Accepted: 52308 Total Submissions: 167414 Difficulty: Easy

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.

這題相當沒有意思，只需要leetcode第102題改一句話就可以。將原來的result.push_back改爲result.insert(result.begin(), temp)就好了，原來是從vector變量的後面增加元素，現在改爲從前面插入元素，那麼自然就倒過來了。

/**
 * Definition for a binary tree node.
 * 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;
vector<int> temp;
vector<TreeNode*> levelNode;
if (root == NULL)
return{};
int count = 0;
int levelNum = 1, nextlevel = 0;
levelNode.push_back(root);
while (!levelNode.empty())
{
for (int i = 0; i < levelNum; i++)
{
temp.push_back(levelNode[0]->val);
if (levelNode[0]->left != NULL)
{
nextlevel++;
levelNode.push_back(levelNode[0]->left);
}
if (levelNode[0]->right != NULL)
{
levelNode.push_back(levelNode[0]->right);
nextlevel++;
}
levelNode.erase(levelNode.begin());
}
result.insert(result.begin(), temp);
temp.clear();
levelNum = nextlevel;
nextlevel = 0;
}
return result;
}
};