Binary Tree Maximum Path Sum

二叉树最大路径和算法

最新推荐文章于 2024-03-25 21:39:45 发布

最新推荐文章于 2024-03-25 21:39:45 发布 · 81 阅读

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本文介绍了一种寻找给定二叉树中最大路径和的算法实现。该路径可以开始并结束于任意节点，通过递归遍历左子树和右子树，计算每个节点的最大贡献值，并维护全局最大路径和。

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 {
    public int maxPathSum(TreeNode root) {
        int[] max = new int[1];
        max[0] = Integer.MIN_VALUE;
        solve(root, max);
        return max[0];
    }

	private int solve(TreeNode root, int[] max) {
		if (root == null) {
			return 0;
		}
		int left = solve(root.left, max);
		int right = solve(root.right, max);
		int ans = Math.max(root.val, Math.max(root.val+left, root.val+right));
		max[0] = Math.max(max[0], Math.max(ans, root.val+left+right));
		return ans;
	}
}