刷题笔记14——树的子结构

题目描述

输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)

分析

要查找树A中是否存在和树B结构一样的子树,可分成两步:
第一步在树A中找到和B的根节点的值一样的结点N,其实就是树的遍历
第二步再判断树A中以N为根结点的子树是不是包含和树B一样的结构

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        bool result = false;
        if (pRoot1 != NULL && pRoot2 != NULL) {
            if(pRoot1->val == pRoot2->val)
                result = DoesTree1HaveTree2(pRoot1, pRoot2);
            if(!result)
                result = HasSubtree(pRoot1->left, pRoot2);
            if(!result)
                result = HasSubtree(pRoot1->right, pRoot2);
        }
        return result;
    }

    bool DoesTree1HaveTree2(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        if (pRoot2 == NULL)
            return true;
        if (pRoot1 == NULL)
            return false;
        if (pRoot1->val != pRoot2->val)
            return false;
        return DoesTree1HaveTree2(pRoot1->left, pRoot2->left) && DoesTree1HaveTree2(pRoot1->right, pRoot2->right);
    }
};

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转载自blog.csdn.net/CSDN_dzh/article/details/84555258