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]
]
简单的bfs而已，不过在bfs的过程中应该注意将相应的数组reverse一下，其实都到最终结果之后在隔行reverse也是可以的，下面给出非递归的版本，用递归同样也很好实现：
/**
* 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 {
struct Node
{
TreeNode * node;
int level;
Node(){}
Node(TreeNode * n, int lv)
: node(n), level(lv){}
};
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> ret;
if(!root) return ret;
vector<int> tmp;
int dep = ;
queue<Node>q;
q.push(Node(root, ));
while(!q.empty()){
Node tmpNode = q.front(); //非递归的使用bfs，借助队列特性
if(tmpNode.node->left)
q.push(Node(tmpNode.node->left, tmpNode.level + ));
if(tmpNode.node->right)
q.push(Node(tmpNode.node->right, tmpNode.level + ));
if(dep < tmpNode.level){
if(dep % ){
reverse(tmp.begin(), tmp.end());
}
ret.push_back(tmp);
tmp.clear();
dep = tmpNode.level;
}
tmp.push_back(tmpNode.node->val);
q.pop();
}
if(dep % ){
reverse(tmp.begin(), tmp.end());
}
ret.push_back(tmp);
return ret;
}
};
java版本的代码如下所示，这里用的是dfs而非bfs，在最后重新reverse就可以了：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ret = new ArrayList<List<Integer>>();
dfs(ret, 1, root);
for(int i = 0; i < ret.size(); ++i){
if(i%2 != 0) Collections.reverse(ret.get(i));
}
return ret;
} void dfs(List<List<Integer>>ret, int dep, TreeNode root){
if(root == null)
return;
if(ret.size() < dep){
List<Integer> list = new ArrayList<Integer>();
list.add(root.val);
ret.add(list);
}else
ret.get(dep - 1).add(root.val);
dfs(ret, dep + 1, root.left);
dfs(ret, dep + 1, root.right);
}
}