剑指 Offer 26. 树的子结构
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));
}
boolean recur(TreeNode A, TreeNode B){
if(B == null) return true; //终止条件
if(A == null || A.val != B.val) return false; //终止条件
return recur(A.left, B.left) && recur(A.recur, B.right);
}
}

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