LeetCode in Kotlin
513. Find Bottom Left Tree Value
Medium
Given the root of a binary tree, return the leftmost value in the last row of the tree.
Example 1:
Input: root = [2,1,3]
Output: 1
Example 2:
Input: root = [1,2,3,4,null,5,6,null,null,7]
Output: 7
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -231 <= Node.val <= 231 - 1
Solution
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
private fun func(root: TreeNode?, level: Int): IntArray? {
if (root!!.left == null && root.right == null) {
val a = IntArray(2)
a[0] = root.`val`
a[1] = level
return a
}
var a: IntArray? = null
var b: IntArray? = null
if (root.left != null) {
a = func(root.left, level + 1)
}
if (root.right != null) {
b = func(root.right, level + 1)
}
return if (a == null) {
b
} else if (b == null) {
a
} else {
if (a[1] >= b[1]) {
a
} else {
b
}
}
}
fun findBottomLeftValue(root: TreeNode?): Int {
val a = func(root, 0)
return if (a != null && a.size > 0) {
a[0]
} else {
-1
}
}
}