[剑指 Offer 第 2 版第 32 题] “从上往下打印二叉树”做题记录

第 32-1 题：不分行从上往下打印二叉树

传送门：不分行从上往下打印二叉树，牛客网 online judge 地址。

从上往下打印出二叉树的每个结点，同一层的结点按照从左到右的顺序打印。

样例：

输入如下图所示二叉树 [8, 12, 2, null, null, 6, null, 4, null, null, null] 8 / \ 12 2 / 6 / 4

输出：[8, 12, 2, 6, 4]

思路：层序遍历，借助利用队列实现。其实就是层序遍历，使用的辅助的数据结构是“队列”，在遍历的过程中，注意到一些细节就可以了。在实现的过程中，要防止：

1、将空元素放入队列； 2、队列元素出队的时候，先判断队列是否为空，非空才可以出队。

Python 代码：

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


    class Solution:
        def printFromTopToBottom(self, root):
            """
            :type root: TreeNode
            :rtype: List[int]
            """
            if root is None:
                return []
            queue = [root]
            res = []
            while queue:
                # 弹出队首元素，索引编号 0 不要忘记写了
                top = queue.pop(0)
                res.append(top.val)
                if top.left:
                    queue.append(top.left)
                if top.right:
                    queue.append(top.right)
            return res

Java 代码：

  import java.util.ArrayList;
    import java.util.LinkedList;

    class TreeNode {
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;

        public TreeNode(int val) {
            this.val = val;
        }
    }

    public class Solution {
        public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
            ArrayList<Integer> res = new ArrayList<>();
            if (root == null) {
                return res;
            }
            LinkedList<TreeNode> queue = new LinkedList<>();
            queue.addLast(root);
            while (!queue.isEmpty()) {
                int curSize = queue.size();
                for (int i = 0; i < curSize; i++) {
                    TreeNode curNode = queue.removeFirst();
                    res.add(curNode.val);
                    if (curNode.left != null) {
                        queue.addLast(curNode.left);
                    }
                    if (curNode.right != null) {
                        queue.addLast(curNode.right);
                    }
                }
            }
            return res;
        }

        public static void main(String[] args) {
            TreeNode node8 = new TreeNode(8);
            TreeNode node6 = new TreeNode(6);
            TreeNode node10 = new TreeNode(10);
            TreeNode node5 = new TreeNode(5);
            TreeNode node7 = new TreeNode(7);
            TreeNode node9 = new TreeNode(9);
            TreeNode node11 = new TreeNode(11);

            node8.left = node6;
            node8.right = node10;

            node6.left = node5;
            node6.right = node7;

            node10.left = node9;
            node10.right = node11;

            Solution solution = new Solution();
            ArrayList<Integer> printFromTopToBottom = solution.PrintFromTopToBottom(node8);
            System.out.println(printFromTopToBottom);
        }
    }

作者：liweiwei1419

来源：https://liweiwei1419.github.io/sword-for-offer/