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

方法一：BFS

为了实现锯齿形层序遍历，需要在层序遍历的基础上增加一个标志位 left，用于标记当前层的节点值的顺序。如果 left 为 true，则当前层的节点值按照从左到右的顺序存入结果数组 ans 中；如果 left 为 false，则当前层的节点值按照从右到左的顺序存入结果数组 ans 中。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点数。

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: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        ans = []
        left = 1
        while q:
            t = []
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t if left else t[::-1])
            left ^= 1
        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) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        boolean left = true;
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.poll();
                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) {
        vector<vector<int>> ans;
        if (!root) return ans;
        queue<TreeNode*> q{{root}};
        int left = 1;
        while (!q.empty()) {
            vector<int> t;
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                t.emplace_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.emplace_back(t);
            left ^= 1;
        }
        return ans;
    }
};

Go

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

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) {
    const ans = [];
    if (!root) {
        return ans;
    }
    const q = [root];
    let left = 1;
    while (q.length) {
        const t = [];
        for (let n = q.length; n; --n) {
            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 ^= 1;
    }
    return ans;
};

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function zigzagLevelOrder(root: TreeNode | null): number[][] {
    const res = [];
    if (root == null) {
        return res;
    }
    let isDesc = false;
    const queue = [root];
    while (queue.length !== 0) {
        const arr = queue.slice().map(() => {
            const { val, left, right } = queue.shift();
            left && queue.push(left);
            right && queue.push(right);
            return val;
        });
        res.push(isDesc ? arr.reverse() : arr);
        isDesc = !isDesc;
    }
    return res;
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
    pub fn zigzag_level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        let mut res = vec![];
        if root.is_none() {
            return res;
        }
        let mut is_desc = false;
        let mut q = VecDeque::new();
        q.push_back(root);
        while !q.is_empty() {
            let mut arr = vec![];
            for _ in 0..q.len() {
                if let Some(node) = q.pop_front().unwrap() {
                    let mut node = node.borrow_mut();
                    arr.push(node.val);
                    if node.left.is_some() {
                        q.push_back(node.left.take());
                    }
                    if node.right.is_some() {
                        q.push_back(node.right.take());
                    }
                }
            }
            if is_desc {
                arr.reverse();
            }
            is_desc = !is_desc;
            res.push(arr);
        }
        res
    }
}

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