124. Binary Tree Maximum Path Sum

[Tcdian]

124. Binary Tree Maximum Path Sum

给定一个 非空 二叉树，返回其最大路径和。

本题中，路径被定义为一条从树中任意节点出发，达到任意节点的序列。该路径 至少包含一个节点，且不一定经过根节点。

Example 1

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

[Tcdian]

Solution ( DFS )

• JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxPathSum = function(root) {
    let result = -Infinity;
    findMaxSinglePath(root);
    return result;

    function findMaxSinglePath(root) {
        if (root === null) {
            return 0;
        }
        const leftPath = findMaxSinglePath(root.left);
        const rightPath = findMaxSinglePath(root.right);
        result = Math.max(Math.max(leftPath, 0) + Math.max(rightPath, 0) + root.val, result);
        return Math.max(leftPath, rightPath, 0) + root.val;
    }
};

• TypeScript Solution

/**
 * 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 result = -Infinity;
    findMaxSinglePath(root);
    return result;

    function findMaxSinglePath(root: TreeNode | null): number {
        if (root === null) {
            return 0;
        }
        const leftPath = findMaxSinglePath(root.left);
        const rightPath = findMaxSinglePath(root.right);
        result = Math.max(Math.max(leftPath, 0) + Math.max(rightPath, 0) + root.val, result);
        return Math.max(leftPath, rightPath, 0) + root.val;
    }
};