算法训练营day33(补)，复习二叉树1

// 889. 根据前序和后序遍历构造二叉树

//   前序中左右 后序遍历左右中

func constructFromPrePost(preorder []int, postorder []int) *TreeNode {

  if len(preorder) == 0 {

    return nil

  }

  root := &TreeNode{}

  root.Val = preorder[0]

  //前序数组去掉root节点

  preorder = preorder[1:]

  //后序数组去掉root节点

  postorder = postorder[:len(postorder)-1]

  //如果前序数组只剩1个，说明只有左子树或者右子树

  if len(postorder) == 1 {

    root.Left = &TreeNode{Val: postorder[0]}

    return root

  }

  index := 0

  for ; index < len(postorder); index++ {

    if postorder[index] == preorder[0] {

      break

    }

  }

  if len(postorder) > 1 {

    leftPostorder := postorder[:index+1]

    rightPostorder := postorder[index+1:]

    leftPreorder := preorder[:index+1]

    rightPreorder := preorder[index+1:]

    root.Left = constructFromPrePost(leftPreorder, leftPostorder)

    root.Right = constructFromPrePost(rightPreorder, rightPostorder)

  }

  return root

}