【LeetCode】103. Binary Tree Zigzag Level Order Traversal
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题目:
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]
]
分析:
这道题的要求是按照Z字形输出一棵二叉树。我的想法是使用BFS来求解这道题。
设两个变量cur和next来记录当前层次和下一层次还未遍历的结点数,设一个bool变量tag来标志当前层应该从左向右输出还是从右向左输出。
按BFS将每一个结点遍历一遍,当cur为0时表示当前层的结点已经全部遍历完,则将该层的结点按这一层应该存储的顺序存储到result中。
这道题的时间复杂度为O(N).
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 {public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector<vector<int>> result; queue<TreeNode*> q; vector<int> vv; if(root == NULL) return result; q.push(root); int cur = 0,next = 0,depth = 0; cur++; bool tag = true; //true表示从左到右,false表示从右到左 while(!q.empty()) { TreeNode* temp = q.front(); vv.push_back(temp->val); cur--; q.pop(); if(temp->left != NULL) { q.push(temp->left); next++; } if(temp->right != NULL) { q.push(temp->right); next++; } if(cur == 0) //当前层已经遍历完了 { if(tag) { result.push_back(vector<int>()); for(int i = 0;i<vv.size();i++) result[depth].push_back(vv[i]); tag = false; } else { result.push_back(vector<int>()); for(int i = vv.size()-1;i>=0;i--) result[depth].push_back(vv[i]); tag = true; } cur = next; next = 0; depth++; vv.clear(); } } return result; }};
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