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: return its zigzag level order traversal as: |
给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。 例如: 返回锯齿形层次遍历如下: |
思路:
第一种:和 上一题很类似,直接 层序遍历,每次遍历到偶数层时翻转就行
/**
* 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) {
if(!root) return {};
vector<vector<int>> res;
queue<TreeNode*> q{{root}};
bool b=false;
while(!q.empty())
{
vector<int> vec;
for(int i =q.size();i>0;--i)
{
TreeNode *p=q.front(); q.pop();
vec.push_back(p->val);
if(p->left) q.push(p->left);
if(p->right) q.push(p->right);
}
if(b) reverse(vec.begin(),vec.end());
b=!b;
res.push_back(vec);
}return res;
}
};
第二种:用两个栈,相邻两个栈分别存到两个栈中,一个从左到右入栈,一个从右到左入栈