2021-02-07：给定两棵二叉树的头节点head1和head2，如何判断head1中是否有某个子树的结构和head2完...

vlambda
2021-02-08

2021-02-07：给定两棵二叉树的头节点head1和head2，如何判断head1中是否有某个子树的结构和head2完...

福哥答案2021-02-07：

对head1和head2序列化为str1和str2。然后用kmp算法去判断str2是否是str1的子串。如果是，head2是子树；如果不是，head2不是子树。

代码用golang编写，代码如下：

package main
import "fmt"
func main() { root := &TreeNode{} root.Val = 1
 root.Left = &TreeNode{} root.Left.Val = 2
 root.Right = &TreeNode{} root.Right.Val = 3
 root.Left.Right = &TreeNode{} root.Left.Right.Val = 4
 root.Right.Left = &TreeNode{} root.Right.Left.Val = 5
 fmt.Println(IsSubTree(root, root.Right))
}
type TreeNode struct { Val int Left *TreeNode Right *TreeNode}
//序列化func serialize(head *TreeNode) string { ansVal := "" ans := &ansVal process(head, ans) return (*ans)[1:]}
func process(head *TreeNode, ans *string) { if head == nil { *ans += ",N" return } *ans += fmt.Sprintf(",%d", head.Val) process(head.Left, ans) //*ans += fmt.Sprintf(",%d", head.Val) process(head.Right, ans) //*ans += fmt.Sprintf(",%d", head.Val)}func getNextArr(m string) []int { mLen := len(m) if mLen == 1 { return []int{-1} } ret := make([]int, mLen) ret[0] = -1 cn := 0 for i := 2; i < mLen; i++ { if m[i] == m[cn] { cn++ ret[i] = cn i++ } else if cn > 0 { cn = ret[cn] } else { ret[i] = 0 i++ } } return ret}
//求子串位置func kmp(s string, m string) int { sLen := len(s) mLen := len(m) if sLen < mLen { return -1 } next := getNextArr(m) x := 0 y := 0 for x < sLen && y < mLen { if s[x] == m[y] { x++ y++ } else if next[y] >= 0 { y = next[y] } else { x++ } } if y == mLen { return x - y } else { return -1 }}
//求是否是子树func IsSubTree(head1 *TreeNode, head2 *TreeNode) bool { if head2 == nil { return true } if head1 == nil { return false } if kmp(serialize(head1), serialize(head2)) >= 0 { return true } else { return false }}