一、问题描述
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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
二、思路
分层遍历二叉树,又叫广度优先遍历二叉树。思路是递归,本题中相当于是分层存储二叉树的节点,必然要涉及到二叉树的深度level。
首先用一个二维数组存储结果,如果当前二维数组的大小不大于当前二叉树的深度,那么必须新建一个数组,将当前的值存入;如果当前二维数组的大小大于二叉树深度,则需要分情况讨论:即看深度是否是2的倍数。
最后分别递归左子树和右子树。
三、代码
/**
* 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 {
private:
void levelOrder(TreeNode* node,vector<vector<int>> &path,int level){
if(!node)
return;
if(path.size() <= level){
vector<int> temp;
temp.push_back(node -> val);
path.push_back(temp);
}else{
if(level % 2 != 0)
path[level].insert(path[level].begin(),node -> val);
else
path[level].push_back(node -> val);
}
levelOrder(node -> left, path, level + 1);
levelOrder(node -> right, path, level + 1);
}
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> path;
if(!root)
return path;
levelOrder(root,path,0);
return path;
}
};