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

Return6.

 

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 & curNo De. 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 ;}}

Recursive solution: For the current node, calculate the max path sum in three cases.

 

Leetcode-Bianry Tree Maximum Path Sum