[LeetCode] 103. 二叉树的锯齿形层次遍历

题目链接 : https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/

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）。

例如：

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:

上一题一样102. 二叉树的层次遍历

思路一:BFS

思路二:递归

代码:

# 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]]:
        if not root: return []
        res = []
        cur_level = [root]
        depth = 0
        while cur_level:
            tmp = []
            next_level = []
            for node in cur_level:
                tmp.append(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            if depth % 2 == 1:
                res.append(tmp[::-1])
            else:
                res.append(tmp)
            depth += 1
            cur_level = next_level
        return res

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<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Deque<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int depth = 0;
        while (!queue.isEmpty()) {
            List<Integer> tmp = new LinkedList<>();
            int cnt = queue.size();
            for (int i = 0; i < cnt; i++) {
                TreeNode node = queue.poll();
                // System.out.println(node.val);
                if (depth % 2 == 0) tmp.add(node.val);
                else tmp.add(0, node.val);
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
            res.add(tmp);
            depth++;
        }
        return res;
    }
}

思路二:

# 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]]:
        res = []

        def helper(root, depth):
            if not root: return
            if len(res) == depth:
                res.append([])
            if depth % 2 == 0:res[depth].append(root.val)
            else: res[depth].insert(0, root.val)
            helper(root.left, depth + 1)
            helper(root.right, depth + 1)
        helper(root, 0)
        return res

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<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        helper(res, root, 0);
        return res;

    }

    private void helper(List<List<Integer>> res, TreeNode root, int depth) {
        if (root == null) return;
        if (res.size() == depth) res.add(new LinkedList<>());
        if (depth % 2 == 0) res.get(depth).add(root.val);
        else res.get(depth).add(0, root.val);
        helper(res, root.left, depth + 1);
        helper(res, root.right, depth + 1);
    }
}