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Leetcode/algorithms/AddOneRowToTree.js at master · superman06/Leetcode · GitHub
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// Source : https://leetcode.com/problems/add-one-row-to-tree
// Author : Dean Shi
// Date : 2017-07-09
/***************************************************************************************
*
* 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 original left subtree should be the left subtree of the new
* left subtree root, its original 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 original tree, and the
* original tree is the new root's left subtree.
*
* Example 1:
*
* Input:
* A binary tree as following:
* 4
* / \
* 2 6
* / \ /
* 3 1 5
*
* v = 1
*
* d = 2
*
* Output:
* 4
* / \
* 1 1
* / \
* 2 6
* / \ /
* 3 1 5
*
* Example 2:
*
* Input:
* A binary tree as following:
* 4
* /
* 2
* / \
* 3 1
*
* v = 1
*
* d = 3
*
* Output:
* 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.
*
*
***************************************************************************************/
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
*
@param
{
TreeNode
} root
*
@param
{
number
} v
*
@param
{
number
} d
*
@return
{
TreeNode
}
*/
var
addOneRow
=
function
(
root
,
v
,
d
,
dir
=
'left'
)
{
if
(
d
===
1
)
{
const
node
=
new
TreeNode
(
v
)
;
[
root
,
node
[
dir
]
]
=
[
node
,
root
]
}
else
if
(
root
)
{
root
.
left
=
addOneRow
(
root
.
left
,
v
,
d
-
1
,
'left'
)
root
.
right
=
addOneRow
(
root
.
right
,
v
,
d
-
1
,
'right'
)
}
return
root
}
;
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