LeetCode in Kotlin
889. Construct Binary Tree from Preorder and Postorder Traversal
Medium
Given two integer arrays, preorder and postorder where preorder is the preorder traversal of a binary tree of distinct values and postorder is the postorder traversal of the same tree, reconstruct and return the binary tree.
If there exist multiple answers, you can return any of them.
Example 1:
Input: preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]
Example 2:
Input: preorder = [1], postorder = [1]
Output: [1]
Constraints:
1 <= preorder.length <= 301 <= preorder[i] <= preorder.length- All the values of
preorderare unique. postorder.length == preorder.length1 <= postorder[i] <= postorder.length- All the values of
postorderare unique. - It is guaranteed that
preorderandpostorderare the preorder traversal and postorder traversal of the same binary tree.
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 {
fun constructFromPrePost(preorder: IntArray, postorder: IntArray): TreeNode? {
return if (preorder.isEmpty() || preorder.size != postorder.size) {
null
} else buildTree(preorder, 0, preorder.size - 1, postorder, 0, postorder.size - 1)
}
private fun buildTree(
preorder: IntArray,
preStart: Int,
preEnd: Int,
postorder: IntArray,
postStart: Int,
postEnd: Int
): TreeNode? {
if (preStart > preEnd || postStart > postEnd) {
return null
}
val data = preorder[preStart]
val root = TreeNode(data)
if (preStart == preEnd) {
return root
}
var offset = postStart
while (offset <= preEnd) {
if (postorder[offset] == preorder[preStart + 1]) {
break
}
offset++
}
root.left = buildTree(
preorder,
preStart + 1,
preStart + offset - postStart + 1,
postorder,
postStart,
offset
)
root.right = buildTree(
preorder,
preStart + offset - postStart + 2,
preEnd,
postorder,
offset + 1,
postEnd - 1
)
return root
}
}