LeetCode OJ: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}

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7]
[9,20],
[3],
]

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

算法思想：

从前往后，从右向左处理完，然后逆序输出

/**
* Definition for binary tree
* 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)return result;
list<list<int>> r;
list<int> t;
int curLev=1;
int nextLev=0;
queue<TreeNode *> que;
que.push(root);
while(!que.empty()){
TreeNode *cur=que.front();
que.pop();
t.push_front(cur->val);
if(cur->right){
nextLev++;
que.push(cur->right);
}
if(cur->left){
nextLev++;
que.push(cur->left);
}
if(--curLev==0){
r.push_front(t);
t.clear();
curLev=nextLev;
nextLev=0;
}
}
for(auto &v:r){
vector<int> k;
for(int i:v){
k.push_back(i);
}
result.push_back(k);
}
return result;
}
};

递归版

/**
* Definition for binary tree
* 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;
traverse(root,1,result);
reverse(result.begin(),result.end());
return result;
}
void traverse(TreeNode *root,size_t level,vector<vector<int>> &result){
if(!root)return;
if(level>result.size())
result.push_back(vector<int>());
result[level-1].push_back(root->val);
traverse(root->left,level+1,result);
traverse(root->right,level+1,result);
}
};