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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

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]:
        if root is None:
            return []
        q = deque()
        q.append(root)
        res = []
        while q:
            size = len(q)
            for _ in range(size):
                node = q.popleft()
                res.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return res

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<>();
        List<Integer> t = new ArrayList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            while (size-- > 0) {
                TreeNode node = q.poll();
                t.add(node.val);
                if (node.left != null) q.offer(node.left);
                if (node.right != null) q.offer(node.right);
            }
        }
        int i = 0, n = t.size();
        int[] res = new int[n];
        for (Integer e : t) {
            res[i++] = e;
        }
        return res;
    }
}

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) {
    if (!root) return [];
    let queue = [root];
    let res = [];
    while (queue.length) {
        let node = queue.shift();
        if (!node) continue;
        res.push(node.val);
        queue.push(node.left, node.right);
    }
    return res;
};

Go

func levelOrder(root *TreeNode) []int {
	if root == nil {
		return []int{}
	}
	q := []*TreeNode{}
	q = append(q, root)
	// 层序遍历,用队列,遍历到谁,就把谁的左右结点加入队列
	res := []int{}
	for len(q) != 0 {
		tmp := q[0]
		q = q[1:]
		res = append(res, tmp.Val)
		if tmp.Left != nil {
			q = append(q, tmp.Left)
		}
		if tmp.Right != nil {
			q = append(q, tmp.Right)
		}
	}
	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<int> levelOrder(TreeNode* root) {
        vector<int> res;
        queue<TreeNode*> q;
        if (root != nullptr) {
            q.push(root);
        }
        while (!q.empty()) {
            TreeNode* node = q.front();
            q.pop();
            if (node->left != nullptr) {
                q.push(node->left);
            }
            if (node->right != nullptr) {
                q.push(node->right);
            }
            res.push_back(node->val);
        }
        return res;
    }
};

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 res = [];
    if (root == null) {
        return res;
    }
    const queue = [root];
    while (queue.length !== 0) {
        const { val, left, right } = queue.shift();
        res.push(val);
        left && queue.push(left);
        right && queue.push(right);
    }
    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 level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = Vec::new();
        let mut queue = VecDeque::new();
        if let Some(node) = root {
            queue.push_back(node);
        }
        while let Some(node) = queue.pop_front() {
            let mut node = node.borrow_mut();
            res.push(node.val);
            if let Some(l) = node.left.take() {
                queue.push_back(l);
            }
            if let Some(r) = node.right.take() {
                queue.push_back(r);
            }
        }
        res
    }
}

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();
    }
}

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