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] ]
此题所谓的Zig-Zag 遍历,及第一行从左到有,第二行从右到左,第三行从左到右,依次类推。此题仍然可以在层次遍历后,对结果进行处理。代码如下:
class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root)
{
//同层次遍历
vector<vector<int>>result;
if (!root)
return result;
queue<TreeNode*> vec;
vec.push(root);
while (vec.size() > 0) {
queue<TreeNode*> tmp_vec;
vector<int> tmp_result;
while (vec.size() > 0) {
TreeNode* node = vec.front();
vec.pop();
if (node->left)
tmp_vec.push(node->left);
if (node->right)
tmp_vec.push(node->right);
tmp_result.push_back(node->val);
}
vec = tmp_vec;
result.push_back(tmp_result);
}
//对第2层、4层。。。进行翻转
for (int i = 1; i < result.size(); i += 2) {
reverse(result[i].begin(), result[i].end());
}
return result;
}
};

本文介绍了一种特殊的二叉树遍历方法——Zig-Zag遍历,并提供了一个C++实现示例。该遍历方式按层级进行,奇数层从左至右,偶数层从右至左。
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