124 Binary Tree Maximum Path Sum

124. Binary Tree Maximum Path Sum

1. Question

Given a 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.

For example: Given the below binary tree,

       1
      / \
     2   3

Return6.

2. Implementation

(1) DFS

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

    public int findMaxPathSum(TreeNode node, int[] max) {
        if (node == null) {
            return 0;
        }

        int leftPathSum = Math.max(0, findMaxPathSum(node.left, max));
        int rightPathSum = Math.max(0, findMaxPathSum(node.right, max));

        max[0] = Math.max(max[0], node.val + leftPathSum + rightPathSum);
        return node.val + Math.max(leftPathSum, rightPathSum);
    }
}

3. Time & Space Complexity

(1) DFS: 时间复杂度: O(n), 空间复杂度: O(h)