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LeetCode: Binary Tree Zigzag Level Order Traversal
LeetCode: Binary Tree Zigzag Level Order Traversal
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]]
地址：https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/
算法：变着花样来遍历一颗树，二叉树都快被玩坏了。还是在二叉树层次遍历的结果上进行处理。代码：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 vector<vector<int> > zigzagLevelOrder(TreeNode *root) {13 if(!root) return vector<vector<int> >();14 queue<TreeNode *> que;15 que.push(root);16 int push_num = 1;17 int pop_num = 0;18 vector<vector<int> > result;19 vector<int> temp;20 int last = 1;21 while(!que.empty()){22 TreeNode *pop_node = que.front();23 temp.push_back(pop_node->val);24 que.pop();25 ++pop_num;26 if(pop_node->left){27 que.push(pop_node->left);28 ++push_num;29 }30 if(pop_node->right){31 que.push(pop_node->right);32 ++push_num;33 }34 if(pop_num == last){35 result.push_back(temp);36 temp.clear();37 last = push_num;38 }39 }40 for(int i = 1; i < result.size(); i += 2){41 int len = result[i].size();42 int half_len = len / 2;43 for(int j = 0; j < half_len; ++j){44 int t = result[i][j];45 result[i][j] = result[i][len-j-1];46 result[i][len-j-1] = t;47 }48 }49 return result;50 }51 };
LeetCode: Binary Tree Zigzag Level Order Traversal
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