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.

这题相当没有意思，只需要leetcode第102题改一句话就可以。将原来的result.push_back改为result.insert(result.begin(), temp)就好了，原来是从vector变量的后面增加元素，现在改为从前面插入元素，那么自然就倒过来了。
/**

* 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;

}

};