[LeetCode]572. Subtree of Another Tree

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本文详细介绍了LeetCode上编号为572的问题“另一个树的子树”的解决方法。通过递归地比较两棵树的节点来判断一棵树是否是另一棵树的子树，并提供了完整的C++实现代码。

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[LeetCode]572. Subtree of Another Tree

题目描述

这里写图片描述

思路

对树s的每一个节点，都判断是否与树t是同一棵树
判断同一棵树使用深搜即可

代码

#include <iostream>

using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if (s == NULL)
            return t == NULL;
        return isSameNode(s, t) || isSubtree(s->left, t) || isSubtree(s->right, t);
    }
    bool isSameNode(TreeNode* s, TreeNode* t) {
        if (s == NULL)
            return t == NULL;
        if (t == NULL)
            return false;
        return s->val == t->val && isSameNode(s->left, t->left) && isSameNode(s->right, t->right);
    }
};

int main() {
    TreeNode *n1 = new TreeNode(3);
    TreeNode *n2 = new TreeNode(4);
    TreeNode *n3 = new TreeNode(5);
    TreeNode *n4 = new TreeNode(1);
    TreeNode *n5 = new TreeNode(2);
    TreeNode *n6 = new TreeNode(0);

    n1->left = n2, n1->right = n3;
    n2->left = n4, n2->right = n5;
    n5->left = n6;

    TreeNode *s1 = new TreeNode(4);
    TreeNode *s2 = new TreeNode(1);
    TreeNode *s3 = new TreeNode(2);

    s1->left = s2, s1->right = s3;

    Solution s;
    cout << s.isSubtree(n1, s1) << endl;

    system("pause");
    return 0;
}