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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：

• 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++

class Solution {
public:
    vector<int> levelOrder(TreeNode* root) {
        vector<int> ret;
        if (!root) {
            return ret;
        }

        queue<TreeNode *> q;
        q.push(root);

        while (!q.empty()) {
            auto head = q.front();
            q.pop();
            ret.push_back(head->val);
            if (head->left) {
                q.push(head->left);
            }

            if (head->right) {
                q.push(head->right);
            }
        }

        return ret;
    }
};

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