Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
Given the root of a binary tree, flatten the tree into a “linked list”:
The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
The “linked list” should be in the same order as a pre-order traversal of the binary tree.
There is a family tree rooted at 0 consisting of n nodes numbered 0 to n - 1. You are given a 0-indexed integer array parents, where parents[i] is the parent for node i. Since node 0 is the root, parents[0] == -1.
There are 105 genetic values, each represented by an integer in the inclusive range [1, 105]. You are given a 0-indexed integer array nums, where nums[i] is a distinct genetic value for node i.
Return an array ans of length n where ans[i] isthe smallest genetic value that is missing from the subtree rooted at nodei.
The subtree rooted at a node x contains node x and all of its descendant nodes.
Example 1:
Input: parents = [-1,0,0,2], nums = [1,2,3,4]
Output: [5,1,1,1]
Explanation: The answer for each subtree is calculated as follows:
- 0: The subtree contains nodes [0,1,2,3] with values [1,2,3,4]. 5 is the smallest missing value.
- 1: The subtree contains only node 1 with value 2. 1 is the smallest missing value.
- 2: The subtree contains nodes [2,3] with values [3,4]. 1 is the smallest missing value.
- 3: The subtree contains only node 3 with value 4. 1 is the smallest missing value.
Example 2:
Input: parents = [-1,0,1,0,3,3], nums = [5,4,6,2,1,3]
Output: [7,1,1,4,2,1]
Explanation: The answer for each subtree is calculated as follows:
- 0: The subtree contains nodes [0,1,2,3,4,5] with values [5,4,6,2,1,3]. 7 is the smallest missing value.
- 1: The subtree contains nodes [1,2] with values [4,6]. 1 is the smallest missing value.
- 2: The subtree contains only node 2 with value 6. 1 is the smallest missing value.
- 3: The subtree contains nodes [3,4,5] with values [2,1,3]. 4 is the smallest missing value.
- 4: The subtree contains only node 4 with value 1. 2 is the smallest missing value.
- 5: The subtree contains only node 5 with value 3. 1 is the smallest missing value.
Example 3:
Input: parents = [-1,2,3,0,2,4,1], nums = [2,3,4,5,6,7,8]
Output: [1,1,1,1,1,1,1]
Explanation: The value 1 is missing from all the subtrees.
Constraints:
n == parents.length == nums.length
2 <= n <= 105
0 <= parents[i] <= n - 1 for i != 0
parents[0] == -1
parents represents a valid tree.
1 <= nums[i] <= 105
Each nums[i] is distinct.
Solution: DFS on a single path
One ancestors of node with value of 1 will have missing values greater than 1. We do a dfs on the path that from node with value 1 to the root.
Time complexity: O(n + max(nums)) Space complexity: O(n + max(nums))
There is a binary tree rooted at 0 consisting of n nodes. The nodes are labeled from 0 to n - 1. You are given a 0-indexed integer array parents representing the tree, where parents[i] is the parent of node i. Since node 0 is the root, parents[0] == -1.
Each node has a score. To find the score of a node, consider if the node and the edges connected to it were removed. The tree would become one or more non-empty subtrees. The size of a subtree is the number of the nodes in it. The score of the node is the product of the sizes of all those subtrees.
Return the number of nodes that have the highest score.
Example 1:
Input: parents = [-1,2,0,2,0]
Output: 3
Explanation:
- The score of node 0 is: 3 * 1 = 3
- The score of node 1 is: 4 = 4
- The score of node 2 is: 1 * 1 * 2 = 2
- The score of node 3 is: 4 = 4
- The score of node 4 is: 4 = 4
The highest score is 4, and three nodes (node 1, node 3, and node 4) have the highest score.
Example 2:
Input: parents = [-1,2,0]
Output: 2
Explanation:
- The score of node 0 is: 2 = 2
- The score of node 1 is: 2 = 2
- The score of node 2 is: 1 * 1 = 1
The highest score is 2, and two nodes (node 0 and node 1) have the highest score.
Constraints:
n == parents.length
2 <= n <= 105
parents[0] == -1
0 <= parents[i] <= n - 1 for i != 0
parents represents a valid binary tree.
Solution: Recursion
Write a function that returns the element of a subtree rooted at node.
We can compute the score based on: 1. size of the subtree(s) 2. # of children
Root is a special case whose score is max(c[0], 1) * max(c[1], 1).
There are n cities numbered from 1 to n. You are given an array edges of size n-1, where edges[i] = [ui, vi] represents a bidirectional edge between cities ui and vi. There exists a unique path between each pair of cities. In other words, the cities form a tree.
A subtree is a subset of cities where every city is reachable from every other city in the subset, where the path between each pair passes through only the cities from the subset. Two subtrees are different if there is a city in one subtree that is not present in the other.
For each d from 1 to n-1, find the number of subtrees in which the maximum distance between any two cities in the subtree is equal to d.
Return an array of sizen-1where the dthelement (1-indexed) is the number of subtrees in which the maximum distance between any two cities is equal to d.
Notice that the distance between the two cities is the number of edges in the path between them.
Example 1:
Input: n = 4, edges = [[1,2],[2,3],[2,4]]
Output: [3,4,0]
Explanation:
The subtrees with subsets {1,2}, {2,3} and {2,4} have a max distance of 1.
The subtrees with subsets {1,2,3}, {1,2,4}, {2,3,4} and {1,2,3,4} have a max distance of 2.
No subtree has two nodes where the max distance between them is 3.
Example 2:
Input: n = 2, edges = [[1,2]]
Output: [1]
Example 3:
Input: n = 3, edges = [[1,2],[2,3]]
Output: [2,1]
Constraints:
2 <= n <= 15
edges.length == n-1
edges[i].length == 2
1 <= ui, vi <= n
All pairs (ui, vi) are distinct.
Solution1: Brute Force+ Diameter of tree
Try all subtrees and find the diameter of that subtree (longest distance between any node)
Time complexity: O(2^n * n) Space complexity: O(n)
C++
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// Author: Huahua
classSolution{
public:
vector<int>countSubgraphsForEachDiameter(intn,vector<vector<int>>& edges, int last = -1) {
