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] ]
标准BFS,层序遍历。
/**
* 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>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> res;
vector<int> traversal;
int level = 0;
queue<TreeNode*> que;
if(root==NULL) return res;
que.push(root);
while(!que.empty()){
int len = que.size();
traversal.clear();
for(int i=0; i<len; ++i){//BFS搜索每次把一整层的内容出队列
TreeNode* FNode = que.front();
traversal.push_back(FNode->val);
if(FNode->left) que.push(FNode->left);
if(FNode->right) que.push(FNode->right);
que.pop();
}
if(level%2==0){
res.push_back(traversal);
}
if(level%2==1){
res.push_back(vector<int>(traversal.rbegin(),traversal.rend()));//翻转vector数组
}
level++;
}
return res;
}
};