/**
* 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) {
if(A == null || B == null) {
return false;
}
if(judge(A, B)) {
return true;
}
return isSubStructure(A.left, B) || isSubStructure(A.right, B);
}
boolean judge(TreeNode A, TreeNode B) {
if(A == null || B == null || A.val != B.val) {
return false;
}
boolean result = true;
if(B.left != null) {
result = judge(A.left, B.left);
}
if(B.right != null) {
result &= judge(A.right, B.right);
}
return result;
}
}
Leetcode_剑指Offer26_树的子结构_递归
最新推荐文章于 2025-01-01 00:15:00 发布
本文探讨了一种通过递归方法解决二叉树问题的实例,介绍了如何利用TreeNode类定义节点,并详细解释了isSubStructure方法,用于检查二叉树A是否包含二叉树B的子结构。judge函数负责节点值的匹配和子树的深度比较。

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