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]
]
Solution 1: 普通层次遍历，借助queue
/**
* 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) { //runtime: 8ms
vector<vector<int>> vec;
if(!root)return vec;
vector<int> v; queue<TreeNode*> q1,q2;
q1.push(root);
while(!q1.empty()){
TreeNode* temp = q1.front();
q1.pop();
v.push_back(temp->val);
if (temp->left)
q2.push(temp->left);
if (temp->right)
q2.push(temp->right); if(q1.empty()){
if (!v.empty())
vec.push_back(v);
v.clear();
swap(q1, q2);
}
}
reverse(vec.begin(),vec.end()); //or vector<vector<int>> ret(vec.rbegin(),vec.rend());return ret;
return vec;
}
}
Solution 2: 递归， 待续