Given the root of a binary tree, construct a 0-indexed m x n string matrix res that represents a formatted layout of the tree. The formatted layout matrix should be constructed using the following rules:
The height of the tree is height and the number of rows m should be equal to height + 1.
The number of columns n should be equal to 2height+1 - 1.
Place the root node in the middle of the top row (more formally, at location res[0][(n-1)/2]).
For each node that has been placed in the matrix at position res[r][c], place its left child at res[r+1][c-2height-r-1] and its right child at res[r+1][c+2height-r-1].
Continue this process until all the nodes in the tree have been placed.
Any empty cells should contain the empty string "".
Return the constructed matrix res.
Example 1:
Input: root = [1,2]
Output:
[["","1",""],
["2","",""]]
Example 2:
Input: root = [1,2,3,null,4]
Output:
[["","","","1","","",""],
["","2","","","","3",""],
["","","4","","","",""]]
Constraints:
The number of nodes in the tree is in the range [1, 210].
-99 <= Node.val <= 99
The depth of the tree will be in the range [1, 10].
纯 ArrayList 的实现，不难。
Code
public List<List<String>> printTree(TreeNode root) {
int h = height(root);
List<List<String>> ans = new ArrayList<List<String>>(h);
for (int i = 0; i < h; i++) {
List<String> tmp = new ArrayList<String>();
for (int j = 0; j < Math.pow(2, h) - 1; j++) {
tmp.add("");
}
ans.add(tmp);
}
dfs(ans, root, (int) Math.pow(2, h - 1) - 1, 0, h);
return ans;
}
private void dfs(List<List<String>> ans, TreeNode node, int col, int row, int h) {
if (node == null) return;
List<String> printRow = ans.get(row);
printRow.set(col, "" + node.val);
if (h > row + 1) {
dfs(ans, node.left, col - (int) Math.pow(2, h - row - 2), row + 1, h);
dfs(ans, node.right, col + (int) Math.pow(2, h - row - 2), row + 1, h);
}
}
private int height(TreeNode node) {
if (node == null) return 0;
return 1 + Math.max(height(node.left), height(node.right));
}