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]
]
这个题目要采用广度优先遍历，所以我们需要一个队列，为了记录节点的深度我用了一个map映射来记录节点的深度
当vec的长度小于节点的深度时，就需要插入一个新的向量，否则直接在向量上插入节点的val
最后我们再翻转vec，即为所求答案
/**
* 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:
void breadthFirstSearch(TreeNode* root, vector<vector<int>> &vec)
{
if (!root)return;
queue<TreeNode*> que;
map<TreeNode*, int> ma;
que.push(root);
ma[root] = ;
while (!que.empty())
{
TreeNode* node = que.front();
if (node->left)
{
que.push(node->left);
ma[node->left] = ma[node] + ;
}
if (node->right)
{
que.push(node->right);
ma[node->right] = ma[node] + ;
}
if (vec.size() <= ma[node])
{
vector<int> childvec;
childvec.push_back(node->val);
vec.push_back(childvec);
}
else
{
vec[ma[node]].push_back(node->val);
}
que.pop(); }
}
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> vec;
breadthFirstSearch(root, vec);
reverse(vec.begin(), vec.end());
return vec;
}
};