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 nodecurat the depthdepth - 1, create two tree nodes with valuevalascur'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 == 1that means there is no depthdepth - 1at all, then create a tree node with valuevalas the new root of the whole original tree, and the original tree is the new root's left subtree.
Example 1:
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:
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 <= 10^51 <= depth <= the depth of tree + 1
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
}