107. 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.

题意：二叉树的层序遍历，从底部向上层序遍历。

思路：按正常的从顶至下层序遍历后，逆着存入另一数组即可。参看102. Binary Tree Level Order Traversal
的代码，加上两行即可。

vector<vector<int>> result(m.rbegin(), m.rend());//构造函数的使用

class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> m;
if (root){
vector<TreeNode*> levelTree1;
vector<TreeNode*> levelTree2;
levelTree1.push_back(root);
while (!(levelTree1.empty() && levelTree2.empty())){
vector<int> p;
if (!levelTree1.empty()){
for (int i = 0; i < levelTree1.size(); i++){
TreeNode* t = levelTree1[i];
p.push_back(t->val);
if (t->left)
levelTree2.push_back(t->left);
if (t->right)
levelTree2.push_back(t->right);
}
levelTree1.clear();
}
else{
for (int i = 0; i < levelTree2.size(); i++){
TreeNode* t = levelTree2[i];
p.push_back(t->val);
if (t->left)
levelTree1.push_back(t->left);
if (t->right)
levelTree1.push_back(t->right);
}
levelTree2.clear();
}
m.push_back(p);
}
}
vector<vector<int>> result(m.rbegin(), m.rend());
return result;
}
};