[leetcode]Binary Tree Maximum Path Sum

Cabinathor

于 2015-08-12 16:50:29 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/47447155

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本文介绍了一个解决二叉树最大路径和问题的算法实现。该算法通过递归方式遍历二叉树节点，计算从任意节点开始的最大路径和，并最终返回整个二叉树的最大路径和。

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from : https://leetcode.com/submissions/detail/36066022/

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.

/**
 * 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 = 0;

	public int maxPathSum(TreeNode root) {
		if (null != root) {
			max = root.val;
		}
		sums(root);
		return max;
	}

	private int sums(TreeNode root) {
		if (null == root) {
			return 0;
		}

		int lsum = sums(root.left);
		int rsum = sums(root.right);

		int sum = root.val;
		if (lsum > 0) {
			sum += lsum;
		}
		if (rsum > 0) {
			sum += rsum;
		}

		if (sum > max) {
			max = sum;
		}

		// as a child, return (root.val+left) or (root.val+right)
		int off = 0;
		if (lsum > 0 && rsum > 0) {
			off = lsum < rsum ? lsum : rsum;
		}
		return sum - off;
	}
}