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8.10 [LeetCode] 173 Binary Tree Maximum Path Sum

时间:2015-08-11 07:16:29      阅读:212      评论:0      收藏:0      [点我收藏+]

Qestion

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, (the node.val may < 0)

       1
      /      2   3
/ \
  4 5

Return 11.

 

Analysis

get leftMax, rightMax, then compare the root , root + left, root + right and (left + root + right)

Code

 

public int maxPathSum(TreeNode root) {
        int max = Integer.MIN_VALUE;
        calculateSum(root, max);
        return max;
    }
     
    public int calculateSum(TreeNode root, int max) {
        if (root == null)
            return 0;
        int left = calculateSum(root.left, max);
        int right = calculateSum(root.right, max);
     
        int current = Math.max(root.val, Math.max(root.val + left, root.val + right)); // current path which can be added to another root
     
        max = Math.max(max, Math.max(current, left + root.val + right));
        return current; // Notice here we return the legal path, not max, or we will get the sum of not a path.
    }

 

8.10 [LeetCode] 173 Binary Tree Maximum Path Sum

原文:http://www.cnblogs.com/michael-du/p/4719863.html

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