Skip to content

Latest commit

 

History

History
131 lines (91 loc) · 4.48 KB

binary-tree-maximum-path-sum.md

File metadata and controls

131 lines (91 loc) · 4.48 KB

124. Binary Tree Maximum Path Sum - 二叉树中的最大路径和

Tags - 题目标签

Description - 题目描述

EN:

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 non-empty 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

ZH-CN:

二叉树中的 路径 被定义为一条节点序列,序列中每对相邻节点之间都存在一条边。同一个节点在一条路径序列中 至多出现一次 。该路径 至少包含一个 节点,且不一定经过根节点。

路径和 是路径中各节点值的总和。

给你一个二叉树的根节点 root ,返回其 最大路径和

 

示例 1:

输入:root = [1,2,3]
输出:6
解释:最优路径是 2 -> 1 -> 3 ,路径和为 2 + 1 + 3 = 6

示例 2:

输入:root = [-10,9,20,null,null,15,7]
输出:42
解释:最优路径是 15 -> 20 -> 7 ,路径和为 15 + 20 + 7 = 42

 

提示:

  • 树中节点数目范围是 [1, 3 * 104]
  • -1000 <= Node.val <= 1000

Link - 题目链接

LeetCode - LeetCode-CN

Latest Accepted Submissions - 最近一次 AC 的提交

Language Runtime Memory Submission Time
typescript 72 ms 51.1 MB 2022/09/01 21:09
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function maxPathSum(root: TreeNode | null): number {
  let max = -Infinity;
  function findMaxContribution(node: TreeNode | null): number {
    if (!node) {
      return 0;
    }
    const leftGain = Math.max(findMaxContribution(node.left), 0);
    const rightGain = Math.max(findMaxContribution(node.right), 0);


    const priceNewPath = node.val + leftGain + rightGain;

    max = Math.max(max, priceNewPath);

    return node.val + Math.max(leftGain, rightGain)
  }

  findMaxContribution(root);
  return max;
}

My Notes - 我的笔记

No notes