LeetCode 107: Binary Tree Level Order Traversal II

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}

,



return its bottom-up level order traversal as:

解题思路

在 LeetCode 102 的基础上 reverse 一下就可以了。

/**
* 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:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> result;

if (root == NULL) return result;

queue<TreeNode *> nodeQueue;
nodeQueue.push(root);
while (!nodeQueue.empty()) {
vector<int> level_values;
int level_size = nodeQueue.size(); // Note: 判断每层的元素个数
for (int i = 0; i < level_size; ++i) {
TreeNode *front = nodeQueue.front();
nodeQueue.pop();

level_values.push_back(front->val);

if (front->left != NULL) nodeQueue.push(front->left);
if (front->right != NULL) nodeQueue.push(front->right);
}
result.push_back(level_values);
}
// reverse
reverse(result.begin(), result.end());
return result;
}
};