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

问题描述

从上到下按层打印二叉树，同一层的节点按从左到右的顺序打印，每一层打印到一行。

例如:
给定二叉树: [3,9,20,null,null,15,7],

 3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

[
  [3],
  [9,20],
  [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<>();

        queue.add(root);
        while(!queue.isEmpty()){
            int k = queue.size();
            List<Integer> tmp = new ArrayList<>();
            for(int i = 0; i < k; i++){
                TreeNode t = queue.poll();
                tmp.add(t.val);
                if(t.left != null) queue.add(t.left);
                if(t.right != null) queue.add(t.right);
            }
            res.add(tmp);
        }

        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 = []

        while queue:
            size = len(queue)
            tmp = []
            for i in range(size):
                node = queue.pop(0)
                tmp.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res.append(tmp)

        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) {
        vector<vector<int>> res;
        if (!root) {
            return res;
        }
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            int n = q.size();
            vector<int> level;
            for (int i = 0; i < n; ++i) {
                auto t = q.front();
                q.pop();
                level.push_back(t->val);
                if (t->left) {
                    q.push(t->left);
                }
                if (t->right) {
                    q.push(t->right);
                }
            }
            res.push_back(level);
        }
        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{}
    }
    var res [][]int
    queue := []*TreeNode{root}
    for len(queue) > 0 {
        size := len(queue)
        level := []int{}
        for i := 0; i < size; i++ {
            node := queue[0]
            queue = queue[1:]
            level = append(level, node.Val)
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        res = append(res, level)
    }
    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 = [];

    queue.push(root);
    while (queue.length > 0) {
        var k = queue.length;
        var tmp = [];
        for (var i = 0; i < k; i++) {
            var t = queue.shift();
            tmp.push(t.val);
            if (t.left != null) queue.push(t.left);
            if (t.right != null) queue.push(t.right);
        }
        res.push(tmp);
    }

    return res;
};

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