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 does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

方法：分治策略,動態規劃。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
//  這是我根據網上的例子，自己重寫的
    private int max;
    private int maxSideSum(TreeNode root) {
        if (root == null) return 0;
        int left = maxSideSum(root.left);
        int right = maxSideSum(root.right);
        int v = left + root.val + right;
        if (v > max) max = v;
        int sum = root.val;
        if (root.val+left>sum) sum = root.val+left;
        if (root.val+right>sum) sum = root.val+right;
        if (sum>max) max = sum;
        return sum;
    }
    public int maxPathSum(TreeNode root) {
        if (root == null) return 0;
        max = root.val;
        maxSideSum(root);
        return max;
    }
    
//  這是網上搜索的解決方案，非常簡潔
//  http://www.programcreek.com/2013/02/leetcode-binary-tree-maximum-path-sum-java/
}
 

更簡潔的版本：

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