LeetCode 124. Binary Tree Maximum Path Sum-优快云博客

本文介绍了一种求解二叉树中任意起止点构成的最大路径和问题的算法实现。通过递归方法，考虑路径包含根节点的情况，算法能够高效地找到最大路径和。

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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

Return 6.

求二叉树任意起止点构成的路径的值最大和

虽然是任意起止点，但此路径一定会有根节点，因此对于该根节点：
max = root.val+maxPathSum(左子树)(>0才加)+maxPathSum(右子树)(>0才加);

但对于每次递归的返回，则应该是
return max(root.val,root.val+maxPathSum(左子树),root.val+maxPathSum(右子树));

 public static int max;

            public static int maxPathSum(TreeNode root) {
                if(root==null)
                    return 0;
                max = Integer.MIN_VALUE;
                maxPathSum2(root);
                return max;
            }
            public static int maxPathSum2(TreeNode root) {
                if(root==null)
                    return 0;
                int temp1 = maxPathSum2(root.left);
                int temp2 = maxPathSum2(root.right);
                int temp = root.val;
                if(temp1>0)temp+=temp1;
                if(temp2>0)temp += temp2;
                max = temp>max?temp:max;
                return max(root.val,root.val+temp1,root.val+temp2);

            }
            public static int max(int i,int j,int k){
                return i > (j>k?j:k) ? i : (j>k?j:k);
            }