先固定队列的大小,以此确定每层的个数,将出队元素入栈(先右孩子再左孩子),最后输出即可
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
vector<vector<int>> levelOrderBottom(TreeNode* root)
{
stack<TreeNode*> st; //存储输出节点
deque<TreeNode*> queue;
vector<vector<int>> res;//存储最终结果
vector<int> sizeOfEachLevel; //存储每层节点个数
if (root == NULL) return res;
queue.push_back(root);
/* 层次遍历,按照顺序将节点入栈,并记录每层节点个数 */
while (!queue.empty())
{
int size = queue.size();
sizeOfEachLevel.push_back(size);
for (int i = 0; i < size; i++)
{
TreeNode* cur = queue.front();
queue.pop_front();
st.push(cur);
if (cur->right != NULL)
queue.push_back(cur->right);
if (cur->left != NULL)
queue.push_back(cur->left);
}
}
/* 输出 */
/* 从最后一层倒序输出 */
for (int i = sizeOfEachLevel.size() - 1; i >= 0; i--)
{
vector<int> curLevel; //存储当前层结果
int j = sizeOfEachLevel[i]; //当前层节点总数
for (; j > 0; j--)
{
int top = st.top()->val;
st.pop();
curLevel.push_back(top);
}
res.push_back(curLevel);
}
return res;
}
};