106. Construct Binary Tree from Inorder and Postorder Traversal

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} inorder
 * @param {number[]} postorder
 * @return {TreeNode}
 */
function buildTree(inorder, postorder) {
  if (!postorder.length) {
    return null
  }

  const root = new TreeNode(postorder[postorder.length - 1])
  const index = inorder.indexOf(root.val)
  root.left = buildTree(inorder.slice(0, index), postorder.slice(0, index))
  root.right = buildTree(inorder.slice(index + 1), postorder.slice(index, -1))

  return root
}

Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution(object):

    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if not postorder:
            return None

        root = TreeNode(postorder[-1])
        index = inorder.index(root.val)
        root.left = self.buildTree(inorder[:index], postorder[:index])
        root.right = self.buildTree(inorder[index + 1:], postorder[index:-1])

        return root

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func buildTree(inorder []int, postorder []int) *TreeNode {
	if len(postorder) == 0 {
		return nil
	}

	root := &TreeNode{Val: postorder[len(postorder)-1]}

	idx := 0
	for i, val := range inorder {
		if val == root.Val {
			idx = i
			break
		}
	}

	root.Left = buildTree(inorder[0:idx], postorder[0:idx])
	root.Right = buildTree(inorder[idx+1:], postorder[idx:len(postorder)-1])

	return root
}