剑指 Offer 32 – III. 从上到下打印二叉树-跟着帅地玩转校招，拿捏大厂offer帅地玩Offer

本问题对应的 leetcode 原文链接：剑指 Offer 32 – III. 从上到下打印二叉树

问题描述

请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。

例如:
给定二叉树: [3,9,20,null,null,15,7],

返回其层次遍历结果：

[
  [3],
  [20,9],
  [15,7]
]

提示：

节点总数 <= 1000

解题思路

视频讲解直达：本题视频讲解

代码实现

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if(root == null){
            return new ArrayList<>();
        }

        Queue<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> res = new ArrayList<>();
        int sum = 1;
        queue.add(root);
        while(!queue.isEmpty()){
            int k = queue.size();
            LinkedList<Integer> tmp = new LinkedList<>();
            for(int i = 0; i < k; i++){
                TreeNode t = queue.poll();
                if(sum % 2 = = 1){
                    tmp.add(t.val);
                } else {
                    tmp.addFirst(t.val);
                }

                if(t.left != null) queue.add(t.left);
                if(t.right != null) queue.add(t.right);
            }
            res.add(tmp);
            sum ++;
        }

        return res;
    }
}

时间复杂度：O(n)
额外空间复杂度：容器里最对存放 1/2 的节点，故为 O(n)

Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []

        queue = [root]
        res = []
        sum = 1

        while queue:
            size = len(queue)
            tmp = []
            for i in range(size):
                node = queue.pop(0)
                if sum % 2 == 1:
                    tmp.append(node.val)
                else:
                    tmp.insert(0, node.val)

                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            res.append(tmp)
            sum += 1

        return res

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<vector<int>> levelOrder(TreeNode* root) {
        if (!root) {
            return {};
        }

        queue<TreeNode*> q{{root}};
        vector<vector<int>> res;
        int sum = 1;
        while (!q.empty()) {
            int n = q.size();
            vector<int> level(n);
            for (int i = 0; i < n; ++i) {
                auto t = q.front();
                q.pop();
                if (sum % 2 == 1) {
                    level[i] = t->val;
                } else {
                    level[n - 1 - i] = t->val;
                }

                if (t->left) {
                    q.push(t->left);
                }
                if (t->right) {
                    q.push(t->right);
                }
            }
            res.push_back(level);
            ++sum;
        }
        return res;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) [][]int {
    if root == nil {
        return [][]int{}
    }

    queue := []*TreeNode{root}
    res := [][]int{}
    sum := 1

    for len(queue) > 0 {
        size := len(queue)
        tmp := []int{}
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            if sum % 2 == 1 {
                tmp = append(tmp, node.Val)
            } else {
                tmp = append([]int{node.Val}, tmp...)
            }
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        res = append(res, tmp)
        sum++
    }

    return res
}

JS

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
    if (root == null) {
        return [];
    }

    var queue = [];
    var res = [];
    var sum = 1;

    queue.push(root);
    while (queue.length > 0) {
        var k = queue.length;
        var tmp = [];
        for (var i = 0; i < k; i++) {
            var t = queue.shift();
            if (sum % 2 == 1) {
                tmp.push(t.val);
            } else {
                tmp.unshift(t.val);
            }

            if (t.left != null) queue.push(t.left);
            if (t.right != null) queue.push(t.right);
        }
        res.push(tmp);
        sum++;
    }

    return res;
};

剑指offer题解

// 层序遍历 class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if(root == null){ return res; } Deque<TreeNode> dq = new ArrayDeque<>(); dq.offerLast(root); boolean reverse = false; while(!dq.isEmpty()){ int size = dq.size(); List<Integer> currList = new ArrayList<>(); for(int i = 0; i < size; i++){ TreeNode curr = dq.pollFirst(); currList.add(curr.val); if(curr.left != null){ dq.offerLast(curr.left); } if(curr.right != null){ dq.offerLast(curr.right); } } if(reverse){ Collections.reverse(currList); } reverse = reverse == true ? false : true; res.add(currList); } return res; } }

/** * 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>> levelOrder(TreeNode root) { Queue<TreeNode> queue = new LinkedList<>(); List<List<Integer>> res = new ArrayList<>(); if(root != null) queue.add(root); while(!queue.isEmpty()){ LinkedList<Integer> tmp = new LinkedList<>(); for(int i = queue.size(); i > 0;i--){ TreeNode node = queue.poll(); //偶数层队列头部 if(res.size() % 2 == 0){ tmp.addLast(node.val); } //奇数层 else { tmp.addFirst(node.val); } if(node.left != null) { queue.add(node.left); } if(node.right != null){ queue.add(node.right); } } res.add(tmp); } return res; } }

class Solution { public List<List<Integer>> levelOrder(TreeNode root) { Queue<TreeNode> queue = new LinkedList<>(); List<List<Integer>> ret = new ArrayList<>(); if(root == null) return ret; queue.offer(root); while(!queue.isEmpty()){ List<Integer> list = new ArrayList<>(); int size = queue.size(); for(int i = 0;i < size;i++){ TreeNode cur = queue.poll(); list.add(cur.val); if(cur.left != null) queue.offer(cur.left); if(cur.right != null) queue.offer(cur.right); } ret.add(list); } for(int i=1;i<ret.size();i+=2){ Collections.reverse(ret.get(i)); } return ret; } }

class Solution { public List<List<Integer>> levelOrder(TreeNode root) { Queue<TreeNode> queue = new LinkedList<>(); List<List<Integer>> ret = new ArrayList<>(); if(root == null) return ret; queue.offer(root); int depth = 1; while(!queue.isEmpty()){ LinkedList<Integer> list = new LinkedList<>(); int size = queue.size(); for(int i = 0;i < size;i++){ TreeNode cur = queue.poll(); if(depth % 2 == 1){ //奇数 list.add(cur.val); }else{ //偶数前插 list.addFirst(cur.val); } if(cur.left != null) queue.offer(cur.left); if(cur.right != null) queue.offer(cur.right); } depth++; ret.add(list); } return ret; } }

class Solution { public List<List<Integer>> levelOrder(TreeNode root) { Deque<TreeNode> deque = new LinkedList<>(); List<List<Integer>> ret = new ArrayList<>(); if(root == null) return ret; deque.add(root); while(!deque.isEmpty()){ List<Integer> list = new ArrayList<>(); for(int i = deque.size();i > 0;i--){ TreeNode cur = deque.removeFirst(); list.add(cur.val); //先左后右加入下层节点 if(cur.left != null) deque.addLast(cur.left); if(cur.right != null) deque.addLast(cur.right); } ret.add(list); if(deque.size() == 0) break; list = new ArrayList<>(); for(int i = deque.size();i > 0;i--){ TreeNode cur = deque.removeLast(); list.add(cur.val); // 先右后左加入下层节点 if(cur.right != null) deque.addFirst(cur.right); if(cur.left != null) deque.addFirst(cur.left); } ret.add(list); } return ret; } }

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宁永久会员 2023年3月10日下午4:01

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Gg 普通 2022年12月31日下午11:13

优秀市民普通 2022年11月1日下午4:32

使用双端队列，用res来判断偶数奇数层，偶数层是从尾部出队，奇数层时从头部出队
时间复杂度on

訁訁訁訁訁普通 2022年10月30日下午9:15

使用Collections偶数行反转？(学的)

使用LinkedList特性模拟双端队列

时间复杂度 O(N) ： N 为二叉树的节点数量，即 BFS 需循环 N 次，占用 O(N) ；双端队列的队首和队尾的添加和删除操作的时间复杂度均为 O(1)
空间复杂度 O(N) ：最差情况下，即当树为满二叉树时，最多有 N/2 个树节点同时在 deque 中，使用 O(N) 大小的额外空间。

优化，一次循环奇数偶数层同时进行

奇数层：从左向右打印，先左后右加入下层节点

偶数层：从右向左打印，先右后左加入下层节点

问题描述

解题思路

代码实现

Python

C++

Go

JS

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