由前序和后序构建任一符合的二叉树

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func constructFromPrePost(preorder []int, postorder []int) *TreeNode {
    return build(preorder,0,len(preorder)-1,postorder,0,len(postorder)-1)
}
 
func build(preorder []int,s1,e1 int,postorder []int,s2,e2 int)*TreeNode{
    if s1>e1{
        return nil
    }
    if s1==e1{
        return &TreeNode{preorder[s1],nil,nil}
    }
    //根节点是前序遍历的第一个
    rootVal:=preorder[s1]
    //选择前序遍历的第二个点作为左子树的根节点
    leftVal:=preorder[s1+1]
    //在后序遍历查找左子树根节点的索引
    idx:=s2
    for i:=s2;i<e2;i++{
        if postorder[i]==leftVal{
            idx=i
            break
        }
    }
    //构造根节点
    root:=&TreeNode{}
    root.Val=rootVal
    //递归构造左右节点
    leftSize:=idx-s2+1
    root.Left=build(preorder,s1+1,s1+leftSize,postorder,s2,idx)
    root.Right=build(preorder,s1+leftSize+1,e1,postorder,idx+1,e2-1)
    return root
}