Subestructura de árbol
Enlace de blog de preguntas
https://blog.csdn.net/qq_43349112/article/details/108542248
Enlace de tema
https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof/
descripción
输入两棵二叉树A和B，判断B是不是A的子结构。(约定空树不是任意一个树的子结构)
B是A的子结构， 即 A中有出现和B相同的结构和节点值。
例如:
给定的树 A:
3
/ \
4 5
/ \
1 2
给定的树 B：
4
/
1
返回 true，因为 B 与 A 的一个子树拥有相同的结构和节点值。
限制：
0 <= 节点个数 <= 10000
Ejemplo
Ejemplo 1:
输入：A = [1,2,3], B = [3,1]
输出：false
Ejemplo 2:
输入：A = [3,4,5,1,2], B = [4,1]
输出：true
Plantilla de código inicial
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSubStructure(TreeNode A, TreeNode B) {
}
}
Código
Solución recomendada:
https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof/solution/mian-shi-ti-26-shu-de-zi-jie-gou-xian-xu -bian-li-p /
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSubStructure(TreeNode A, TreeNode B) {
return (A != null && B != null) &&
(recur(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B));
}
private boolean recur(TreeNode p, TreeNode q) {
if (q == null) {
return true;
}
if (p == null || p.val != q.val) {
return false;
}
return recur(p.left, q.left) && recur(p.right, q.right);
}
}