32 - I. 从上到下打印二叉树

---

comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9832%20-%20I.%20%E4%BB%8E%E4%B8%8A%E5%88%B0%E4%B8%8B%E6%89%93%E5%8D%B0%E4%BA%8C%E5%8F%89%E6%A0%91/README.md

面试题 32 - I. 从上到下打印二叉树

题目描述

从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。

 

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回：

[3,9,20,15,7]

 

提示：

1. 节点总数 <= 1000

解法

方法一：BFS

我们可以通过 BFS 遍历二叉树，将每一层的节点值存入数组中，最后返回数组即可。

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, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def levelOrder(self, root: TreeNode) -> List[int]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:#保证层序下去
            for _ in range(len(q)):
                node = q.popleft()
                ans.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return ans

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] levelOrder(TreeNode root) {
        if (root == null) {
            return new int[] {};
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        List<Integer> res = new ArrayList<>();
        while (!q.isEmpty()) {
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.poll();
                res.add(node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        int[] ans = new int[res.size()];
        for (int i = 0; i < ans.length; ++i) {
            ans[i] = res.get(i);
        }
        return ans;
    }
}

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<int> levelOrder(TreeNode* root) {
        if (!root) {
            return {};
        }
        vector<int> ans;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                ans.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        return ans;
    }
};

Go

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

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 levelOrder(root: TreeNode | null): number[] {
    const ans: number[] = [];
    if (!root) {
        return ans;
    }
    const q: TreeNode[] = [root];
    while (q.length) {
        const t: TreeNode[] = [];
        for (const { val, left, right } of q) {
            ans.push(val);
            left && t.push(left);
            right && t.push(right);
        }
        q.splice(0, q.length, ...t);
    }
    return ans;
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option>>,
//   pub right: Option>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut ans = Vec::new();
        let mut q = VecDeque::new();
        if let Some(node) = root {
            q.push_back(node);
        }
        while let Some(node) = q.pop_front() {
            let mut node = node.borrow_mut();
            ans.push(node.val);
            if let Some(l) = node.left.take() {
                q.push_back(l);
            }
            if let Some(r) = node.right.take() {
                q.push_back(r);
            }
        }
        ans
    }
}

JavaScript

/**
 * 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) {
    const ans = [];
    if (!root) {
        return ans;
    }
    const q = [root];
    while (q.length) {
        const t = [];
        for (const { val, left, right } of q) {
            ans.push(val);
            left && t.push(left);
            right && t.push(right);
        }
        q.splice(0, q.length, ...t);
    }
    return ans;
};

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int[] LevelOrder(TreeNode root) {
        if (root == null) {
            return new int[]{};
        }
        Queue<TreeNode> q = new Queue<TreeNode>();
        q.Enqueue(root);
        List<int> ans = new List<int>();
        while (q.Count != 0) {
            int x = q.Count;
            for (int i = 0; i < x; i++) {
                TreeNode node = q.Dequeue();
                ans.Add(node.val);
                if (node.left != null) {
                    q.Enqueue(node.left);
                }
                if (node.right != null) {
                    q.Enqueue(node.right);
                }
            }
        }
        return ans.ToArray();
    }
}

Swift

/* public class TreeNode {
*     var val: Int
*     var left: TreeNode?
*     var right: TreeNode?
*     init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
* }
*/

class Solution {
    func levelOrder(_ root: TreeNode?) -> [Int] {
        guard let root = root else {
            return []
        }

        var queue: [TreeNode] = [root]
        var result: [Int] = []

        while !queue.isEmpty {
            let node = queue.removeFirst()
            result.append(node.val)

            if let left = node.left {
                queue.append(left)
            }
            if let right = node.right {
                queue.append(right)
            }
        }

        return result
    }
}