LeetCode 572 Subtree of Another Tree
2017-08-05 10:33
519 查看
题目:
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 scould also be considered as a subtree of itself.
Example 1:
Given tree s:
Given tree t:
Return true,
because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:
Given tree t:
Return false.
题目链接
题意:
给两个非空的二叉树s和t,编写函数实现判断t是否为s的子树。其中,s也可以是它自己的子树。
先序遍历二叉树s,当判断s当前结点的val和t的val相同,则进入判断子树的函数,同步遍历s和t,看是否完全相同,假如不相同,则进行下一层遍历。
代码如下:
class Solution {
public:
bool dfsSubtree(TreeNode* s, TreeNode * t) { // 判断两个子树是否完全相同
if ((!s && t) || (s && !t)) return false;
else if (!s && !t) return true;
return s->val == t->val && dfsSubtree(s->left, t->left) && dfsSubtree(s->right, t->right);
}
bool isSubtree(TreeNode* s, TreeNode* t) {
if ((!s && t) || (s && !t)) return false;
return ((s->val == t->val) && dfsSubtree(s, t) || isSubtree(s->left, t) || isSubtree(s->right, t)); // 先序遍历同时判断s->val和t->val
}
};
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 scould also be considered as a subtree of itself.
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.
题目链接
题意:
给两个非空的二叉树s和t,编写函数实现判断t是否为s的子树。其中,s也可以是它自己的子树。
先序遍历二叉树s,当判断s当前结点的val和t的val相同,则进入判断子树的函数,同步遍历s和t,看是否完全相同,假如不相同,则进行下一层遍历。
代码如下:
class Solution {
public:
bool dfsSubtree(TreeNode* s, TreeNode * t) { // 判断两个子树是否完全相同
if ((!s && t) || (s && !t)) return false;
else if (!s && !t) return true;
return s->val == t->val && dfsSubtree(s->left, t->left) && dfsSubtree(s->right, t->right);
}
bool isSubtree(TreeNode* s, TreeNode* t) {
if ((!s && t) || (s && !t)) return false;
return ((s->val == t->val) && dfsSubtree(s, t) || isSubtree(s->left, t) || isSubtree(s->right, t)); // 先序遍历同时判断s->val和t->val
}
};
相关文章推荐
- 【Leetcode-easy-572】Subtree of Another Tree
- LeetCode - 572 - Subtree of Another Tree
- leetcode 572 Subtree of Another Tree
- leetcode 572 Subtree of Another Tree
- LeetCode@Tree_572_Subtree_of_Another_Tree
- leetcode(572):Subtree of Another Tree
- LeetCode 572: Subtree of Another Tree
- Leetcode算法学习日志-572 Subtree of Another Tree
- LeetCode: Subtree of Another Tree
- LeetCode Subtree of Another Tree
- [LeetCode] Subtree of Another Tree
- LeetCode Subtree of Another Tree
- leetcode[Subtree of Another Tree]//待整理多种解法
- leetcode : subtree of another tree
- 【LeetCode】Subtree of Another Tree 解题报告
- [LeetCode] Subtree of Another Tree 另一个树的子树
- leetcode专题—Subtree of Another Tree
- Subtree of Another Tree
- Subtree of Another Tree:判断一棵树是否是另一棵树的子树
- Subtree of Another Tree