Leetcode-Bianry 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.

Solution:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

 //NOTE: Need to consider about negtive number, or ask interviewer about this issue!
 //NOTE2: at every node, we need consider about three cases.
 //1. the path start from some node in the lower level and end at the current node, called singlePath.
 //2. the path from some child node in the left and end at some child node at right, called combinePath.
 //3. the path that does not contain the current node, called lowPath.
 //curNode:
 //singlePath = max(left.singlePath, right.singlePath, curNode.val);
 //combinePath = curNode.val+left.singlePath+right.singlePath;
 //lowPath = max(left.combinePath, left.singlePath, left.lowPath, right.ALLPATH);
 //Return:
 //max(root.singlePath, root.combinePath, root.lowPath);
class PathInfo{
    public int singlePath;
    public int combinePath;
    public int lowPath;
    public int singleNodePath;

    public PathInfo(){
        singlePath = 0;
        combinePath = 0;
        lowPath = 0;
    }
}

public class Solution {
    public int maxPathSum(TreeNode root) {
        PathInfo rootInfo = new PathInfo();
        rootInfo = maxPathSumRecur(root);

        int max = rootInfo.singlePath;
        if (rootInfo.combinePath>max)
            max = rootInfo.combinePath;
        if (rootInfo.lowPath>max)
            max = rootInfo.lowPath;

        return max;
    }

    public PathInfo maxPathSumRecur(TreeNode curNode){
        //If current node is a leaf node
        if (curNode.left==null&&curNode.right==null){
            PathInfo path = new PathInfo();
            path.singlePath = curNode.val;
            path.combinePath = curNode.val;
            path.lowPath = curNode.val;
            return path;
        }

        //If not, then get the PathInfo of its child nodes.
        PathInfo left = null;
        PathInfo right = null;
        PathInfo cur = new PathInfo();
        if (curNode.left!=null)
            left = maxPathSumRecur(curNode.left);
        if (curNode.right!=null)
            right = maxPathSumRecur(curNode.right);

        //Now calculate the PathInfo of current node.
        if (right==null)
            cur.singlePath = curNode.val+left.singlePath;
        else if (left==null)
            cur.singlePath = curNode.val+right.singlePath;
        else {
            if (left.singlePath>right.singlePath)
                cur.singlePath = curNode.val+left.singlePath;
            else
                cur.singlePath = curNode.val+right.singlePath;
        }
        if (cur.singlePath<curNode.val)
            cur.singlePath=curNode.val;

        if (right==null)
            cur.combinePath = curNode.val+left.singlePath;
        else if (left==null)
            cur.combinePath = curNode.val+right.singlePath;
        else
            cur.combinePath = curNode.val+left.singlePath+right.singlePath;

        int max = Integer.MIN_VALUE;
        if (right==null){
            max = left.lowPath;
            if (left.combinePath>max)
                max = left.combinePath;
        } else if (left==null){
            max = right.lowPath;
            if (right.combinePath>max)
                max = right.combinePath;
        } else {
            max = left.lowPath;
            if (left.combinePath>max)
                max = left.combinePath;
            if (right.lowPath>max)
                max = right.lowPath;
            if (right.combinePath>max)
                max = right.combinePath;
        }
        if (max<cur.singlePath)
            max=cur.singlePath;

        cur.lowPath = max;

        return cur;
    }
}

递归求解：对于当前node，计算三种情况的max path sum.