124. 二叉树中的最大路径和

class Solution {
    private int maxSum = Integer.MIN_VALUE;
    
    public int maxPathSum(TreeNode root) {
        maxGain(root);
        return maxSum;
    }
    
    private int maxGain(TreeNode node) {
        if (node == null) {
            return 0;
        }
        
        // 递归计算左右子树的最大贡献值
        // 只有当贡献值大于0时，才会选择该子树
        int leftGain = Math.max(maxGain(node.left), 0);
        int rightGain = Math.max(maxGain(node.right), 0);
        
        // 计算当前节点的最大路径和
        // 路径是：左子树 -> 当前节点 -> 右子树
        int currentPathSum = node.val + leftGain + rightGain;
        
        // 更新全局最大路径和
        maxSum = Math.max(maxSum, currentPathSum);
        
        // 返回当前节点对父节点的最大贡献值
        // 只能选择左子树或右子树中的一条路径
        return node.val + Math.max(leftGain, rightGain);
    }
}

// 二叉树节点定义
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {}
    TreeNode(int val) { this.val = val; }
    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}