LeetCode OJ:Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:

Given binary tree

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

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

confused what

"{1,#,2,3}"

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

算法思想：

按照层序遍历思想，根据访问到奇偶数层处理先后问题

/**
* 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> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int>> result;
if(!root)return result;
list<list<int>> r;
list<int> t;
int fang=0;
int curLev=1;
int nextLev=0;
queue<TreeNode *> que;
que.push(root);
while(!que.empty()){
TreeNode *cur=que.front();
que.pop();
if(!fang)
t.push_back(cur->val);
else t.push_front(cur->val);
if(cur->left){
nextLev++;
que.push(cur->left);
}
if(cur->right){
nextLev++;
que.push(cur->right);
}
if(--curLev==0){
r.push_back(t);
t.clear();
curLev=nextLev;
nextLev=0;
fang^=1;
}
}
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> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int>> result;
traverse(root,1,result,true);
return result;
}
void traverse(TreeNode *root,size_t level,vector<vector<int>> &result,bool left_to_right){
if(!root)return;

if(level>result.size())
result.push_back(vector<int>());

if(left_to_right)
result[level-1].push_back(root->val);
else
result[level-1].insert(result[level-1].begin(),root->val);

traverse(root->left,level+1,result,!left_to_right);
traverse(root->right,level+1,result,!left_to_right);
}
};