124. Binary Tree Maximum Path Sum

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本文介绍了一种寻找二叉树中最大路径和的有效算法。路径定义为从任意节点开始到任意节点结束的一系列边，至少包含一个节点。算法采用自底向上的递归方式，通过遍历左右子树来确定每个节点的最大贡献值。

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

自底向顶遍历二叉树，分别遍历左、右子树，结合左、右子树的父亲节点，可以得出当前节点向下求和的最大值，程序如下所示：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int maxVal = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        pathSum(root); 
        return maxVal;
    }
    public int pathSum(TreeNode root) {
        if (root == null){
            return 0;
        }
        int leftSum = Math.max(0, pathSum(root.left));
        int rightSum = Math.max(0, pathSum(root.right));
        maxVal = Math.max(maxVal, leftSum + rightSum + root.val);
        return Math.max(leftSum, rightSum) + root.val;
    }
}