103. 二叉树的锯齿形层序遍历

English Version

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

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
        if root is None:
            return []
        left, ans = False, []
        q = deque([root])
        while q:
            n = len(q)
            t = []
            for _ in range(n):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            if left:
                t.reverse()
            ans.append(t)
            left = not left
        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 Collections.emptyList();
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        boolean left = false;
        List<List<Integer>> ans = new ArrayList<>();
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int i = 0, n = q.size(); i < n; ++i) {
                TreeNode node = q.pollFirst();
                t.add(node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            if (left) {
                Collections.reverse(t);
            }
            ans.add(t);
            left = !left;
        }
        return ans;
    }
}

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) return {};
        queue<TreeNode*> q{{root}};
        vector<vector<int>> ans;
        bool left = false;
        while (!q.empty())
        {
            vector<int> t;
            for (int i = 0, n = q.size(); i < n; ++i)
            {
                auto node = q.front();
                q.pop();
                t.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            if (left) reverse(t.begin(), t.end());
            ans.push_back(t);
            left = !left;
        }
        return ans;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func zigzagLevelOrder(root *TreeNode) [][]int {
	if root == nil {
		return nil
	}
	var ans [][]int
	var q = []*TreeNode{root}
	left := false
	for len(q) > 0 {
		var t []int
		n := len(q)
		for i := 0; i < n; i++ {
			node := q[0]
			q = q[1:]
			t = append(t, node.Val)
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
		if left {
			i, j := 0, n-1
			for i < j {
				t[i], t[j] = t[j], t[i]
				i++
				j--
			}
		}
		ans = append(ans, t)
		left = !left
	}
	return ans
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function (root) {
    if (!root) {
        return [];
    }
    let ans = [];
    let q = [root];
    let left = false;
    while (q.length) {
        let t = [];
        for (let i = 0, n = q.length; i < n; ++i) {
            const node = q.shift();
            t.push(node.val);
            if (node.left) {
                q.push(node.left);
            }
            if (node.right) {
                q.push(node.right);
            }
        }
        if (left) {
            t.reverse();
        }
        ans.push(t);
        left = !left;
    }
    return ans;
};

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