标签：right return OFF26 recur 子结构 && TreeNode null
/遍历A树，用每个节点开始与B进行比较
public boolean isSubStructure(TreeNode A, TreeNode B) {
return (A != null && B != null) && (recur(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B));
}
boolean recur(TreeNode A, TreeNode B) {
if(B == null) return true;
if(A == null || A.val != B.val) return false;
return recur(A.left, B.left) && recur(A.right, B.right);
}
标签：right,
return,
OFF26,
recur,
子结构,
&&,
TreeNode,
null
From： https://www.cnblogs.com/lwx11111/p/16778816.html