二叉树打印

题目
有一棵二叉树，请设计一个算法，按照层次打印这棵二叉树。
给定二叉树的根结点root，请返回打印结果，结果按照每一层一个数组进行储存，所有数组的顺序按照层数从上往下，且每一层的数组内元素按照从左往右排列。保证结点数小于等于500。
我的提交：

-- coding:utf-8 --

class TreeNode:

def init(self, x):

self.val = x

self.left = None

self.right = None

class TreePrinter:
def printTree(self, root):

write code here

    result = []
    queue = []
    if root == None:
        return result
    result.append([root.val])
    queue.append([root])
    while True:
        line = []
        qline = []
        parents = queue[len(result) - 1]
        for i in range(len(parents)):
            node = parents[i]
            if node.left != None:
                qline.append(node.left)
                line.append(node.left.val)
            if node.right != None:
                qline.append(node.right)
                line.append(node.right.val)

        if len(line) == 0:
            return result
        queue.append(qline)
        result.append(line)

参考答案：
/
struct TreeNode {
int val;
struct TreeNode left;
struct TreeNode right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};/

class TreePrinter {
public:
vector > printTree(TreeNode root) {
int dep[550],head = 0,tail = 0;
TreeNode now,*q[550];
q[tail ++] = root; dep[root->val] = 0;
while(head != tail){
now = q[head ++];
if(now->left) q[tail ++] = now->left,dep[now->left->val] = dep[now->val] + 1;
if(now->right) q[tail ++] = now->right,dep[now->right->val] = dep[now->val] + 1;
}
vector > ret;
for(int j,i = 0;i < tail;i = j){
vector tmp;
j = i;
while(j < tail && dep[q[j]->val] == dep[q[i]->val]){
tmp.push_back(q[j]->val);
++ j;
}
ret.push_back(tmp);
}
return ret;
}
};

来源：51CTO

作者：跌底

链接：https://blog.51cto.com/13545923/2054474