一棵树的高度( height of a tree )等于它的根的高。对于图中的树,E的深度为1而高为2;F的深度为1而高也是1;该树的高为3。一棵树的深度( depth of a tree )等于其最深的树叶的深度,该深度总是等于这棵树的高。 如果存在从n1到n2的一条路径,那么n1是n2的一位祖先( ancestor ),而n2是n1的一个后...
树是一个无向图,其中任何两个顶点只通过一条路径连接。 换句话说,任何一个没有简单环路的连通图都是一棵树。 给你一棵包含n个节点的树,标记为0到n - 1。给定数字n和一个有n - 1条无向边的edges列表(每一个边都是一对标签),其中edges[i] = [ai, bi]表示树中节点ai和bi之间存在一条无向边。 可...
https://leetcode.com/problems/minimum-height-trees/ 有trick 对graph不熟悉。再看几遍。 参考http://bookshadow.com/weblog/2015/11/26/leetcode-minimum-height-trees/ 基本思路是“逐层删去叶子节点,直到剩下根节点为止” 有点类似于拓扑排序 最终剩下的节点个数可能为1或者2 时间复杂度:O(n),其中n为...
left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { if(t1==NULL&&t2==NULL)return NULL; if(t1==NULL) return t2; if(t2==NULL) return t1; TreeNode *tmp=(TreeNode *)malloc(sizeof(TreeNode));...
The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels. FormatThe graph contains n nodes which are ...
characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root ...
力扣leetcode-cn.com/problems/diameter-of-binary-tree/ 题目描述 给定一棵二叉树,你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。 示例: 给定二叉树 1 / \ 2 3 / \ 4 5 返回3, 它的长度是路径 [4,2,1,3] 或者 [5...
The height of a node is the length of the longest downward path to a leaf from that node. The height of the root is the height of the tree. The depth of a node is the length of the path to its root (i.e., its root path). This is commonly needed in the manipulation of the ...
The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called min...LeetCode 310. Minimum Height Trees 解法 LeetCode 310. Minimum Height Trees 解法 最简单的解法就是DFS遍历,即从每个节点出发,计算到达末端节点的最大路径长度,那么此路径长度最小的...
**/publicclassTreeNode{int val;//值TreeNode left;//左子树TreeNode right;//右子树TreeNode(){}TreeNode(int val){this.val=val;}TreeNode(int val,TreeNode left,TreeNode right){this.val=val;this.left=left;this.right=right;}} 复制