Breadth-first Search -- Leetcode problem107. Binary Tree Level Order Traversal II

描述：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]

]

分析：层序遍历二叉树，将遍历结果从底层到高层输出

思路一：直接用reverse函数对Binary Tree Level Order Traversal中的vector结果进行反转操作。

class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> my_vec;
vector<int> result;
if (!root) return my_vec;
queue<TreeNode*> my_queue;
my_queue.push(root);
while (!my_queue.empty()) {
int n = my_queue.size();
result.clear();
for (int i = 0; i < n; i ++) {
TreeNode* temp = my_queue.front();
my_queue.pop();
result.push_back(temp -> val);
if (temp -> left) my_queue.push(temp -> left);
if (temp -> right) my_queue.push(temp -> right);
}
my_vec.push_back(result);
}
reverse(my_vec.begin(), my_vec.end());
return my_vec;
}
};