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Copy path103.binary-tree-zigzag-level-order-traversal.cpp
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103.binary-tree-zigzag-level-order-traversal.cpp
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/*
* @lc app=leetcode id=103 lang=cpp
*
* [103] Binary Tree Zigzag Level Order Traversal
*
* https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/
*
* algorithms
* Medium (49.67%)
* Likes: 3088
* Dislikes: 122
* Total Accepted: 474.1K
* Total Submissions: 947.8K
* Testcase Example: '[3,9,20,null,null,15,7]'
*
* 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]
* ]
*
*
*/
// @lc code=start
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
if (root == nullptr)
return result;
queue<TreeNode*> nodeQue;
nodeQue.push(root);
bool leftToRight = true;
while (!nodeQue.empty()) {
int size = nodeQue.size();
vector<int> curLevel(size);
for (int i = 0; i < size; ++i) {
TreeNode* node = nodeQue.front();
nodeQue.pop();
int index = leftToRight? i : size - i - 1;
curLevel[index] = node->val;
if (node->left)
nodeQue.push(node->left);
if (node->right)
nodeQue.push(node->right);
}
leftToRight = !leftToRight;
if (size)
result.push_back(curLevel);
}
return result;
}
};
// @lc code=end