Leetcode 310 - Min height trees

Note:

  • What kind of node should be chosen as our root?
  • Nodes with indegree == 1 can be seen as leaves. Those are def not what we want.
  • Why? Image a path from one leaf from another leaf is like l1 - a - b - c - l2. Here b is the center and distance(b, l1) < distance(l1, l2), distance(b, l2) < distance(l1, l2).
  • If we pick any interior node, its distance to any leaf node will be shorter than picking a leaf node as the root.
  • The more inside a node is, the shorter the path is to our leaf nodes.
  • Use a map to store edges.
  • Use queue to store current leaf nodes.
  • Every time we pop out a leaf and minus 1 from its adjacent nodes. If its neighbor’s indegree === 1, add them to the queue.
  • We need a res var to store the result. In each iteration of BFS, we clear res first, then we add popped nodes into res.
  • After the last round of iteration, the result will be in res.

Question:

A tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.

Given a tree of n nodes labelled from 0 to n - 1, and an array of n - 1`` edges where edges[i] = [ai, bi] indicates that there is an undirected edge between the two nodes ai and bi in the tree, you can choose any node of the tree as the root. When you select a node x as the root, the result tree has height h. Among all possible rooted trees, those with minimum height (i.e. min(h))  are called minimum height trees (MHTs).

Return a list of all MHTs’ root labels. You can return the answer in any order.

The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

Example:

img

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Input: n = 4, edges = [[1,0],[1,2],[1,3]]
Output: [1]
Explanation: As shown, the height of the tree is 1 when the root is the node with label 1 which is the only MHT.

Code:

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/**
* @param {number} n
* @param {number[][]} edges
* @return {number[]}
*/
var findMinHeightTrees = function (n, edges) {
if (n === 1) return [0];
let indegree = [...Array(n).fill(0)];
let map = new Map();
for (let i = 0; i < edges.length; i++) {
const a = edges[i][0];
const b = edges[i][1];
indegree[a]++;
indegree[b]++;

if (!map.has(a)) {
map.set(a, [b]);
} else {
map.set(a, [...map.get(a), b]);
}
if (!map.has(b)) {
map.set(b, [a]);
} else {
map.set(b, [...map.get(b), a]);
}
}
let queue = [];
for (let i = 0; i < indegree.length; i++) {
if (indegree[i] === 1) queue.push(i);
}
let res = [];
while (queue.length > 0) {
// Reset res every time.
res = [];
const length = queue.length;
for (let i = 0; i < length; i++) {
const node = queue.shift();
res.push(node);
indegree[node]--;
const adjacentNodes = map.get(node);
for (const n of adjacentNodes) {
indegree[n]--;
if (indegree[n] === 1) {
queue.push(n);
}
}
}
}
return res;
};