103 Binary Tree Zigzag Level Order Traversal 二叉树的锯齿形层次遍历
Description:
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).
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]
]
题目描述:
给定一个二叉树，返回其节点值的锯齿形层次遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
示例 :
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回锯齿形层次遍历如下：
[
[3],
[20,9],
[15,7]
]
思路:
直接层序遍历, 每两层加上一次翻转, 或者用层数判断之后再考虑加在表头还是表尾
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* 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> zigzagLevelOrder(TreeNode* root)
{
vector> result;
helper(root, 0, result);
return result;
}
private:
void helper(TreeNode* root, int depth, vector> &result)
{
if (!root) return;
if (depth >= result.size()) result.push_back(vector{});
if (depth & 1) result[depth].insert(result[depth].begin(), root -> val);
else result[depth].push_back(root -> val);
helper(root -> left, depth + 1, result);
helper(root -> right, depth + 1, result);
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List> zigzagLevelOrder(TreeNode root) {
List> result = new ArrayList<>();
helper(root, result, 0);
return result;
}
private void helper(TreeNode root, List> result, int depth) {
if (root == null) return;
if (result.size() == depth) result.add(new ArrayList());
if ((depth & 1) == 1) result.get(depth).add(0, root.val);
else result.get(depth).add(root.val);
helper(root.left, result, depth + 1);
helper(root.right, result, depth + 1);
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
def helper(root: TreeNode, depth: int, result: List[List[int]]) -> None:
if not root:
return
if depth >= len(result):
result.append([])
result[depth].append(root.val)
helper(root.left, depth + 1, result)
helper(root.right, depth + 1, result)
result = []
helper(root, 0, result)
for i, item in enumerate(result):
if (i & 1):
item.reverse()
return result