// 树中结点含有分叉,树B是树A的子结构 // 8 8 // / \ / \ // 8 7 9 2 // / \ // 9 2 // / \ // 4 7 void Test1() { BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8); BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8); BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode...
【LeetCode】572. Subtree of Another Tree Given two non-empty binary treessandt, check whether treethas exactly the same structure and node values with a subtree ofs. A subtree ofsis a tree consists of a node insand all of this node’s descendants. The treescould also be considered as a...
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. E...
Leetcode 543. Diameter of Binary Tree 题目描述:找出二叉树的最长直径(也就是所以节点连成最长的路径长度) 题目链接:Leetcode 543. Diameter of Binary Tree 题目思路:用到的就是经典的递归返回值的形式,不断返回左子树和右子树的深度,然后过程中更新最长的路径。 代码如下 参考链接 543. Diameter of Binary Tr...
Given tree t: 4 / \ 1 2 1. 2. 3. 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 1. 2. 3. 4. 5. 6. 7. Given tree t: ...
leetcode 572. Subtree of Another Tree 子树判断 + 深度优先遍历DFS,Giventwonon-emptybinarytreessandt,checkwhethertreethasexactlythesamest
572. Subtree of Another Tree(另一个树的子树) 题目地址:https://leetcode.com/problems/subtree-of-another-tree/description/ Given two non-empty binary treessandt, check whether treethas exactly the same structure and node values with a subtree ofs. A subtree ofsis a tree consists of a node...
Can you solve this real interview question? Subtree of Another Tree - Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise. A subtree of a
//leetcode.com/problems/subtree-of-another-tree/discuss/102741/Python-Straightforward-with-Explanation-(O(ST)-and-O(S+T)-approaches) # Merkle hashing的方法暂时先不了解,调用了库函数面试也可能不让用 def isSubtree(self, s, t): """ :type s: TreeNode :type t: TreeNode :rtype: bool ""...
# https://leetcode.com/problems/subtree-of-another-tree/solution/ def isSubtree(self, s, t): """ :type s: TreeNode :type t: TreeNode :rtype: bool """ s_res = self.preorder(s, True) t_res = self.preorder(t, True) return t_res in s_res def preorder(self, root, isLeft...