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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
解題思路：
本題和102. Binary Tree Level Order Traversal類似俘侠，只是遍歷的時候102. Binary Tree Level Order Traversal是從上到下一層一層遍歷蔗蹋，而本題是從下到上一層層遍歷则吟，基本思路相同甚疟，唯一的區(qū)別是要先求出二叉樹的深度。
具體代碼如下：
class Solution {
public:
int getDepth(TreeNode* root)
{
if(!root) return 0;
return max(getDepth(root->left), getDepth(root->right)) + 1;
}
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> ret;
if(!root) return ret;
int depth = getDepth(root);
ret.resize(depth);
int levelNumber = 0;
queue<TreeNode*> level;
level.push(root);
while(!level.empty())
{
int size = level.size();
for(int i = 0; i < size; ++i)
{
TreeNode* temp = level.front();
level.pop();
ret[depth - levelNumber - 1].push_back(temp->val);
if(temp->left) level.push(temp->left);
if(temp->right) level.push(temp->right);
}
levelNumber ++;
}
return ret;
}
};
