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| 1 | +# 2385. Amount of Time for Binary Tree to Be Infected |
| 2 | + |
| 3 | +- Difficulty: Medium. |
| 4 | +- Related Topics: Tree, Depth-First Search, Breadth-First Search, Binary Tree. |
| 5 | +- Similar Questions: Maximum Depth of Binary Tree, Shortest Path to Get Food, All Nodes Distance K in Binary Tree, Count the Number of Infection Sequences. |
| 6 | + |
| 7 | +## Problem |
| 8 | + |
| 9 | +You are given the `root` of a binary tree with **unique** values, and an integer `start`. At minute `0`, an **infection** starts from the node with value `start`. |
| 10 | + |
| 11 | +Each minute, a node becomes infected if: |
| 12 | + |
| 13 | + |
| 14 | + |
| 15 | +- The node is currently uninfected. |
| 16 | + |
| 17 | +- The node is adjacent to an infected node. |
| 18 | + |
| 19 | + |
| 20 | +Return **the number of minutes needed for the entire tree to be infected.** |
| 21 | + |
| 22 | + |
| 23 | +Example 1: |
| 24 | + |
| 25 | + |
| 26 | + |
| 27 | +``` |
| 28 | +Input: root = [1,5,3,null,4,10,6,9,2], start = 3 |
| 29 | +Output: 4 |
| 30 | +Explanation: The following nodes are infected during: |
| 31 | +- Minute 0: Node 3 |
| 32 | +- Minute 1: Nodes 1, 10 and 6 |
| 33 | +- Minute 2: Node 5 |
| 34 | +- Minute 3: Node 4 |
| 35 | +- Minute 4: Nodes 9 and 2 |
| 36 | +It takes 4 minutes for the whole tree to be infected so we return 4. |
| 37 | +``` |
| 38 | + |
| 39 | +Example 2: |
| 40 | + |
| 41 | + |
| 42 | + |
| 43 | +``` |
| 44 | +Input: root = [1], start = 1 |
| 45 | +Output: 0 |
| 46 | +Explanation: At minute 0, the only node in the tree is infected so we return 0. |
| 47 | +``` |
| 48 | + |
| 49 | + |
| 50 | +**Constraints:** |
| 51 | + |
| 52 | + |
| 53 | + |
| 54 | +- The number of nodes in the tree is in the range `[1, 105]`. |
| 55 | + |
| 56 | +- `1 <= Node.val <= 105` |
| 57 | + |
| 58 | +- Each node has a **unique** value. |
| 59 | + |
| 60 | +- A node with a value of `start` exists in the tree. |
| 61 | + |
| 62 | + |
| 63 | + |
| 64 | +## Solution |
| 65 | + |
| 66 | +```javascript |
| 67 | +/** |
| 68 | + * Definition for a binary tree node. |
| 69 | + * function TreeNode(val, left, right) { |
| 70 | + * this.val = (val===undefined ? 0 : val) |
| 71 | + * this.left = (left===undefined ? null : left) |
| 72 | + * this.right = (right===undefined ? null : right) |
| 73 | + * } |
| 74 | + */ |
| 75 | +/** |
| 76 | + * @param {TreeNode} root |
| 77 | + * @param {number} start |
| 78 | + * @return {number} |
| 79 | + */ |
| 80 | +var amountOfTime = function(root, start) { |
| 81 | + const startNode = findStartNode(root, start, null); |
| 82 | + return findLongestPath(root, startNode, 0, {}); |
| 83 | +}; |
| 84 | + |
| 85 | +var findStartNode = function(node, start, parent) { |
| 86 | + if (!node) return null; |
| 87 | + const startNode = (node.val === start ? node : null) |
| 88 | + || findStartNode(node.left, start, node) |
| 89 | + || findStartNode(node.right, start, node); |
| 90 | + if (startNode) { |
| 91 | + node.parent = parent; |
| 92 | + return startNode; |
| 93 | + } |
| 94 | + return null; |
| 95 | +}; |
| 96 | + |
| 97 | +var findLongestPath = function(root, node, depth, visited) { |
| 98 | + if (!node || visited[node.val]) return 0; |
| 99 | + visited[node.val] = true; |
| 100 | + if (!node.left && !node.right && !node.parent) return depth; |
| 101 | + return Math.max( |
| 102 | + node === root ? depth : 0, |
| 103 | + findLongestPath(root, node.left, depth + 1, visited), |
| 104 | + findLongestPath(root, node.right, depth + 1, visited), |
| 105 | + findLongestPath(root, node.parent, depth + 1, visited), |
| 106 | + ); |
| 107 | +}; |
| 108 | +``` |
| 109 | + |
| 110 | +**Explain:** |
| 111 | + |
| 112 | +nope. |
| 113 | + |
| 114 | +**Complexity:** |
| 115 | + |
| 116 | +* Time complexity : O(n). |
| 117 | +* Space complexity : O(n). |
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