LeetCode

二叉树的锯齿层序遍历

103. 二叉树的锯齿形层序遍历 - 力扣（LeetCode）

题目描述

给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（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 1：

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]

示例 2：

输入：root = [1]
输出：[[1]]

示例 3：

输入：root = []
输出：[]

提示：

• 树中节点数目在范围 [0, 2000] 内
• -100 <= Node.val <= 100

思路

103. 二叉树的锯齿形层序遍历 - 力扣（LeetCode）

代码

C++

/*** 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) {if(root == nullptr){return {};}vector<vector<int>> ans;vector<TreeNode *> cur = {root};for(bool even = false; !cur.empty(); even = !even){vector<TreeNode *> nxt;vector<int> vals;for(auto node : cur){vals.push_back(node->val);if(node->left) nxt.push_back(node->left);if(node->right) nxt.push_back(node->right);}cur = move(nxt);if(even) reverse(vals.begin(),vals.end());ans.emplace_back(vals);}return ans;}
};

Java

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<List<Integer>> zigzagLevelOrder(TreeNode root) {if(root == null){return List.of();}List<List<Integer>> ans = new ArrayList<>();List<TreeNode> cur = new ArrayList<>();cur.add(root);for(boolean even = false; !cur.isEmpty(); even = !even){List<TreeNode> nxt = new ArrayList<>();List<Integer> vals = new ArrayList<>(cur.size());for(TreeNode node : cur){vals.add(node.val);if(node.left != null) nxt.add(node.left);if(node.right != null) nxt.add(node.right);}cur = nxt; if(even) Collections.reverse(vals);ans.add(vals);}return ans;}
}