[LeetCode] 124. Binary Tree Maximum Path Sum

轉自LeetCode

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node's values in the path.

Given the root of a binary tree, return the maximum path sum of any path.

 

Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

 

Constraints:

• The number of nodes in the tree is in the range [1, 3 * 104].
• -1000 <= Node.val <= 1000

Solution

歷遍求極值，然後資料結構是tree，應該會是用 DFS 來解

這題麻煩的應該是條件的限制，因為最大值的 path，不一定會發生在 root 的地方

也可以想成，每個 node 都可以是最大值發生的 root node

所以在 DFS 的過程中，一直去紀錄最大值是多少

然後有個要注意的是，每個 DFS 的回傳值，必須從 left 和 right 之中選一個

原因是題目規定，同一個 path 只能出現一次

如果 left 和 right 都選了，則他們兩個人的 parent node 就會出現兩次，不符合規定

Kotlin

	/**
	* Example:
	* var ti = TreeNode(5)
	* var v = ti.`val`
	* Definition for a binary tree node.
	* class TreeNode(var `val`: Int) {
	* var left: TreeNode? = null
	* var right: TreeNode? = null
	* }
	*/
	class Solution {
	var ans = Int.MIN_VALUE
	fun maxPathSum(root: TreeNode?): Int {
	dfs(root)
	return ans
	}
	fun dfs(node: TreeNode?): Int {
	if (node == null) {
	return 0
	}
	val left = Math.max(0, dfs(node!!.left))
	val right = Math.max(0, dfs(node!!.right))
	ans = Math.max(ans, node!!.`val` + left + right)
	return Math.max(left, right) + node!!.`val` // the path can't be peated so we can only pick left or right
	}
	}

view raw binary_tree_maximum_path_sum.kt hosted with ❤ by GitHub