中序周遊+前序/後序周遊,可以唯一确定一個二叉樹
本題後序周遊從右向左是樹中從右向左的節點
通過後序周遊确定節點的值,在中序周遊中該節點位置可以确定左右子樹節點值
通過中序周遊中确定的節點位置,知道左右子樹節點個數,因為中序周遊數組個數和後序周遊數組個數相同,進而确定後序周遊數組
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(inorder []int, postorder []int) *TreeNode {
if len(inorder)==0 {
return nil
}
head := &TreeNode{Val: postorder[len(postorder)-1], Left: nil, Right: nil}
//用中序的個數判斷後續的個數
lnum := getid(inorder, head.Val)
//rnum := len(inorder)-lnum-1
head.Left = buildTree(inorder[:getid(inorder, head.Val)], postorder[:lnum])
head.Right = buildTree(inorder[getid(inorder, head.Val)+1:], postorder[lnum:len(postorder)-1])
return head
}
func getid(list []int, val int) int{
for i, v := range list{
if v == val{
return i
}
}
return -1
}