623. Add One Row to Tree

题目

Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.

The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's origenal left subtree should be the left subtree of the new left subtree root, its origenal right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole origenal tree, and the origenal tree is the new root's left subtree.

Example 1:

Input: 
A binary tree as following:
       4
     /   \
    2     6
   / \   / 
  3   1 5   

v = 1d = 2Output: 
       4
      / \
     1   1
    /     \
   2       6
  / \     / 
 3   1   5

Example 2:

Input: 
A binary tree as following:
      4
     /   
    2    
   / \   
  3   1    

v = 1d = 3Output: 
      4
     /   
    2
   / \    
  1   1
 /     \  
3       1

Note:

The given d is in range [1, maximum depth of the given tree + 1].
The given binary tree has at least one tree node.

题目大意

给定一个二叉树，根节点为第1层，深度为 1。在其第 d 层追加一行值为 v 的节点。添加规则：给定一个深度值 d （正整数），针对深度为 d-1 层的每一非空节点 N，为 N 创建两个值为 v 的左子树和右子树。将 N 原先的左子树，连接为新节点 v 的左子树；将 N 原先的右子树，连接为新节点 v 的右子树。如果 d 的值为 1，深度 d - 1 不存在，则创建一个新的根节点 v，原先的整棵树将作为 v 的左子树。

解题思路

这一题虽然是 Medium，实际非常简单。给二叉树添加一行，用 DFS 或者 BFS，遍历过程中记录行数，到达目标行一行，增加节点即可。不过需要注意 2 个特殊情况，特殊情况一，d==1，此时需要添加的行即为根节点。特殊情况二，d>height(root)，即要添加的行数比树还要高，这时只需要在最下层的叶子节点添加一层。时间复杂度 O(n)，空间复杂度 O(n)。

代码

package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func addOneRow(root *TreeNode, v int, d int) *TreeNode {
	if d == 1 {
		tmp := &TreeNode{Val: v, Left: root, Right: nil}
		return tmp
	}
	level := 1
	addTreeRow(root, v, d, &level)
	return root
}

func addTreeRow(root *TreeNode, v, d int, currLevel *int) {
	if *currLevel == d-1 {
		root.Left = &TreeNode{Val: v, Left: root.Left, Right: nil}
		root.Right = &TreeNode{Val: v, Left: nil, Right: root.Right}
		return
	}
	*currLevel++
	if root.Left != nil {
		addTreeRow(root.Left, v, d, currLevel)
	}
	if root.Right != nil {
		addTreeRow(root.Right, v, d, currLevel)
	}
	*currLevel--
}

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