Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      / 
     2   3

Return 6.

注意： The path may start and end at any node in the tree. 而不是从叶子到叶子， 所以有很多种情况。

根本的考虑是递归， tree的结构首先考虑递归。

这里从最下面一层开始一层层往上走，直到root。查看每一层的max的值。

每一个root有两边，left and right， 分别查这两边的最值，然后比较，返回的是经过这个node的路径的最大值。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int max;
    public int maxPathSum(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        max = root.val;
        if(root.right == null && root.left == null) return root.val;
        PathForNode(root);
        return max;
    }
    public int PathForNode(TreeNode n){
        if(n == null) return 0;
        else{
            int right = PathForNode(n.right);
            int left = PathForNode(n.left);
            if(max < n.val + right + left) max = n.val + right + left;
            else if(max < n.val + left) max = n.val + left;
            else if(max < n.val + right) max = n.val + right;
            else if(max < n.val) max = n.val;
            return max(n.val + right , n.val + left,n.val);
            }
    }
    public int max(int a,int b,int c){
        if(a < 0 && b < 0 && c < 0) return 0;
        int s = (a < b) ? b : a;
        return (s < c) ? c : s;
    }
}

 第二遍：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int result = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result = Integer.MIN_VALUE;
        dfs(root);
        return result;
    }
    public int dfs(TreeNode root){
        if(root == null) return 0;
        int left = dfs(root.left);
        int right = dfs(root.right);
        result = Math.max(result, left + right + root.val);
        return Math.max(0, Math.max(left, right) + root.val);// 不返回小于0的数字！！！
    }
}