Content

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node's values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

The number of nodes in the tree is in the range [1, 3 * 10⁴].
-1000 <= Node.val <= 1000

Solution

1. 动态规划 + DFS

Java

class Solution {
    int max = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        // The number of nodes in the tree is in the range [1, 3 * 10⁴]
        // -1000 <= Node.val <= 1000
        dfs(0, root);
        return max;
    }
    /**
     * 深度优先搜素
     *
     * @param sum  当前结点的父结点作为path的最后一个端点，path的最大sum
     * @param node 当前结点
     * @return 以当前结点作为起始端点的最大长度
     */
    private int dfs(int sum, TreeNode node) {
        if (null == node) {
            return 0;
        }
        sum = sum <= 0 ? node.val : sum + node.val;
        max = Math.max(max, sum);
        int left = Math.max(0, dfs(sum, node.left));
        int right = Math.max(0, dfs(sum, node.right));
        // 在以当前结点为顶点的子树中，经过当前结点的最大值
        int max0 = node.val + left + right;
        max = Math.max(max, max0);
        return node.val + Math.max(left, right);
    }
}

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {
    }
    TreeNode(int val) {
        this.val = val;
    }
    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}