[LeetCode]572. Subtree of Another Tree(判断树t是否是树S的子树)
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572. Subtree of Another Tree
原题链接
Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.
题目大意:
给定两个非空二进制树s和t,检查树t是否具有与s子树完全相同的结构和节点值。 s的子树是由s中的节点和所有这个节点的后代组成的树。 树也可以被认为是一个本身的子树。
Example 1:
Given tree s:
3 / \ 4 5 / \ 1 2
Given tree t:
4 / \ 1 2
Return true, because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:
3 / \ 4 5 / \ 1 2 / 0
Given tree t:
4 / \ 1 2
Return false.
思路:
- 树s和树t均非空,判断s是不是整体与t相同,相同返回TRUE,不同的话判断s的左子树和右子树是否与t相同,依次递归下去
- 关键是写出一个函数,给定两个二叉树根节点,判断两个二叉树是否相等
- 需要注意的还有代码中注释部分
代码如下:
C++ 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) { //下面这个if很重要,题目给定两个均为非空二叉树,假如s(或s的子树)为空,直接返回false) if (s == nullptr) //非空二叉树 return false; if(isSame(s, t)) return true; return isSubtree(s->left, t) || isSubtree(s->right, t); } //判断两个二叉树是否相同 bool isSame(TreeNode* s, TreeNode* t){ if(s == nullptr && t == nullptr)//节点均为空 return true; if(s == nullptr || t == nullptr)//(节点不是均为空情况下)s为空或者t为空 return false; if(s->val != t->val) return false; return isSame(s->left, t->left) && isSame(s->right, t->right); }};
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