124. Binary Tree Maximum Path Sum

最新推荐文章于 2020-06-11 01:14:01 发布

BigFatSheep

最新推荐文章于 2020-06-11 01:14:01 发布

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分类专栏： Leetcode Hard

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/83210441

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本文探讨了在非空二叉树中寻找最大路径和的问题，通过递归算法结合贪心策略实现，无需路径一定经过根节点。文章详细解释了算法思路及其实现过程。

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Description

Given a non-empty 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 must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

   1
  / \
 2   3

Output: 6
Example 2:

Input: [-10,9,20,null,null,15,7]

-10
/
9 20
/
15 7

Output: 42

Problem URL

Solution

给一棵非空二叉树，找到一条拥有最大加和的路径，并返回它的和。路径不一定要经过root。

We could use recursion to solve this proble. Using a global variable to store the max path sum. Then from the root node, we start recursion, for a node in the tree, if it is null, return 0; Or calculate the maxpath from it left child and right child, If the value is smaller than 0, we just discard it. Calculating new max path value. And return the biggst one from its left and right plus the value of this node for upper level recursion.

Code

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