# 655. Print Binary Tree 输出二叉树

@TOC

## # 题目描述

Print a binary tree in an m*n 2D string array following these rules:

• The row number m should be equal to the height of the given binary tree.
• The column number n should always be an odd number.
• The root node's value (in string format) should be put in the exactly middle of the first row it can be put. The column and the row where the root node belongs will separate the rest space into two parts (left-bottom part and right-bottom part). You should print the left subtree in the left-bottom part and print the right subtree in the right-bottom part. The left-bottom part and the right-bottom part should have the same size. Even if one subtree is none while the other is not, you don't need to print anything for the none subtree but still need to leave the space as large as that for the other subtree. However, if two subtrees are none, then you don't need to leave space for both of them.
• Each unused space should contain an empty string "".
• Print the subtrees following the same rules.

Example 1:

``````Input:
1
/
2
Output:
[["", "1", ""],
["2", "", ""]]
``````

Example 2:

``````Input:
1
/ \
2   3
\
4
Output:
[["", "", "", "1", "", "", ""],
["", "2", "", "", "", "3", ""],
["", "", "4", "", "", "", ""]]
``````

Example 3:

``````Input:
1
/ \
2   5
/
3
/
4
Output:

[["",  "",  "", "",  "", "", "", "1", "",  "",  "",  "",  "", "", ""]
["",  "",  "", "2", "", "", "", "",  "",  "",  "",  "5", "", "", ""]
["",  "3", "", "",  "", "", "", "",  "",  "",  "",  "",  "", "", ""]
["4", "",  "", "",  "", "", "", "",  "",  "",  "",  "",  "", "", ""]]
``````

Note: The height of binary tree is in the range of [1, 10].

## # 解题方法

### # DFS

1. https://leetcode.com/problems/print-binary-tree/discuss/106273/Simple-Python-with-thorough-explanation

2. https://leetcode.com/problems/print-binary-tree/discuss/106250/Python-Straight-Forward-Solution

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

class Solution(object):
def printTree(self, root):
"""
:type root: TreeNode
:rtype: List[List[str]]
"""
if not root: return [""]
def getDepth(root):
if not root:
return 0
return 1 + max(getDepth(root.left), getDepth(root.right))
d = getDepth(root)
cols = 2 ** d - 1
self.res = [["" for i in range(cols)] for j in range(d)]
def helper(root, d, pos):
self.res[-d - 1][pos] = str(root.val)
if root.left: helper(root.left, d - 1, pos - 2 ** (d - 1))
if root.right: helper(root.right, d - 1, pos + 2 ** (d - 1))
helper(root, d - 1, 2 ** (d - 1) - 1)
return self.res
``````

``````/**
* 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<vector<string>> printTree(TreeNode* root) {
const int h = depth(root);
int w = pow(2, h) - 1;
vector<vector<string>> res(h, vector<string>(w, ""));
dfs(root, res, 0, 0, w);
return res;
}
private:
int depth(TreeNode* root) {
if (!root) return 0;
return max(depth(root->left), depth(root->right)) + 1;
}
// [left, right)
void dfs(TreeNode* root, vector<vector<string>>& res, int depth, int left, int right) {
if (!root || depth == res.size()) return;
int mid = left + (right - left) / 2;
res[depth][mid] = to_string(root->val);
dfs(root->left, res, depth + 1, left, mid);
dfs(root->right, res, depth + 1, mid + 1, right);
}
};
``````

### # BFS

C++代码如下：

``````/**
* 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<vector<string>> printTree(TreeNode* root) {
const int h = depth(root);
int w = pow(2, h) - 1;
int curH = -1;
vector<vector<string>> res(h, vector<string>(w, ""));
queue<TreeNode*> q;
q.push(root);
// [first, second)
queue<pair<int, int>> idxQ;
idxQ.push({0, w});
while (!q.empty()) {
curH++;
int size = q.size();
for (int i = 0; i < size; i++) {
TreeNode* node = q.front(); q.pop();
auto idx = idxQ.front(); idxQ.pop();
if (!node) continue;
int left = idx.first, right = idx.second;
int mid = left + (right - left) / 2;
res[curH][mid] = to_string(node->val);
q.push(node->left);
q.push(node->right);
idxQ.push({left, mid});
idxQ.push({mid + 1, right});
}
}
return res;
}
private:
int depth(TreeNode* root) {
if (!root) return 0;
return max(depth(root->left), depth(root->right)) + 1;
}
};
``````

C++代码如下：

``````/**
* 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<vector<string>> printTree(TreeNode* root) {
int d = depth(root);
int w = (1 << d) - 1;
queue<pair<TreeNode*, pair<int, int>>> q; // TreeNode*, start, end
q.push({root, {0, w}});
vector<vector<string>> res(d, vector<string>(w));
int curd = 0;
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; ++i) {
auto f = q.front(); q.pop();
TreeNode* node = f.first;
int start = f.second.first;
int end = f.second.second;
if (!node) continue;
int mid = start + (end - start) / 2;
res[curd][mid] = to_string(node->val);
q.push({node->left, {start, mid - 1}});
q.push({node->right, {mid + 1, end}});
}
++curd;
}
return res;
}
int depth(TreeNode* root) {
if (!root) return 0;
return max(depth(root->left), depth(root->right)) + 1;
}
};
``````

## # 日期

2018 年 3 月 4 日 2018 年 12 月 18 日 —— 改革开放40周年 2019 年 2 月 25 日 —— 二月就要完了