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.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.

思路：先序遍历，用递归，每一层用level记录着，根据level所在的vector插入值即可。

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int getTreeHeight(TreeNode *root){
if(root == NULL)
return 0;
int leftDepth = getTreeHeight(root->left)+1;
int rightDepth = getTreeHeight(root->right)+1;
return max(leftDepth, rightDepth);
}

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

vector<vector<int> > levelOrderBottom(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int treeHeight = getTreeHeight(root);
vector<vector<int> > result(treeHeight);
travelButtom(root, result, 0);
return result;
}
};