[LeetCode] Binary Tree Level Order Traversal II

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.

解题思路：

这道题题意是从下往上逐层遍历二叉树，将结果存在一个二维向量中。可以先求出该树的高度，确定二维向量第一维的大小，然后深度遍历数，将值插入到相应的位置中即可。下面的方法只需8ms。另外题目中给出的问题：{1，#，2，3}表示按数组的形式存储二叉树，#表示对应的节点为空，下标能够表示两个节点的相互关系。

/**
* 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) {
int h = getHeight(root);
vector<vector<int>> result(h, vector<int>());
traversal(root, 1, h, result);
return result;
}

private:

void traversal(TreeNode* root, int level, int h, vector<vector<int>>& result){
if(root!=NULL){
result[h-level].push_back(root->val);
traversal(root->left, level+1, h, result);
traversal(root->right, level+1, h, result);
}
}

int getHeight(TreeNode* root){
if(root==NULL){
return 0;
}
return 1+max(getHeight(root->left), getHeight(root->right));
}
};