一、題目

  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", "", ""]]

二、題目大意

  以二維陣列的方式列印二叉樹。

三、解題思路

  這道題可以算是二叉樹分層遍歷的一個進階題目，難點在於節點如何放在陣列中的合適位置，從上述例子中，可以發現子樹的根節點應該處於當前子樹空間的中心位置：

  ["", "", ""] 1 => (0 + 2) / 2  => ["", "1", ""]
  
  而2則為1的左子樹的根節點

  [""] 2 => (0 + 0) / 2 => ["2"]

四、程式碼實現

const printTree = root => {

  const height = getTreeHeight(root)
  const width = (1 << height) - 1

  // 初始化整個陣列
  const ans = []
  for (let i = 0; i < height; i++) {
    ans[i] = []
    for (let j = 0; j < width; j++) {
      ans[i].push("")
    }
  }

  // 二分填充
  print(root, ans, 0, 0, width - 1)
  return ans
  function getTreeHeight (root) {
    if (!root) {
      return 0
    }
    return Math.max(getTreeHeight(root.left), getTreeHeight(root.right)) + 1
  }

  function print (root, ans, height, start, end) {
    if (!root) {
      return
    }
    const mid = (start + end) / 2
    ans[height][mid] = String(root.val)
    print(root.left, ans, height + 1, start, mid - 1)
    print(root.right, ans, height + 1, mid + 1, end)
  }

}
 

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