623. Add One Row to Tree

Medium
Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.
Note that the root node is at depth 1.
The adding rule is:
  • Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur's left subtree root and right subtree root.
  • cur's original left subtree should be the left subtree of the new left subtree root.
  • cur's original right subtree should be the right subtree of the new right subtree root.
  • If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root's left subtree.
Example 1:
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Input: root = [4,2,6,3,1,5], val = 1, depth = 2
Output: [4,1,1,2,null,null,6,3,1,5]
Example 2:
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Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]
Constraints:
  • The number of nodes in the tree is in the range [1, 10^4].
  • The depth of the tree is in the range [1, 10^4].
  • -100 <= Node.val <= 100
  • -10^5 <= val <= 10^5
  • 1 <= depth <= the depth of tree + 1

解題

Runtime: 2 ms, faster than 100%
Memory Usage: 5.6 MB, less than 50%
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
if root == nil { return nil }
​
if depth == 1 {
return &TreeNode{ Val: val, Left: root}
}
if depth == 2 {
newLeft := &TreeNode{ Val: val, Left: root.Left }
newRight := &TreeNode{ Val: val, Right: root.Right }
return &TreeNode{ Val: root.Val, Left: newLeft, Right: newRight }
}
​
root.Left, root.Right = addOneRow(root.Left, val, depth-1), addOneRow(root.Right, val, depth-1)
​
return root
}