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  • LeetCode: Binary Tree Maximum Path Sum 解题报告

    Binary Tree Maximum Path Sum
    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

    SOLUTION 1:

    计算树的最长path有2种情况:

    1. 通过根的path.

      (1)如果左子树从左树根到任何一个Node的path大于零,可以链到root上

      (2)如果右子树从右树根到任何一个Node的path大于零,可以链到root上

    2. 不通过根的path. 这个可以取左子树及右子树的path的最大值。

    所以创建一个inner class:

    记录2个值:

    1. 本树的最大path。

    2. 本树从根节点出发到任何一个节点的最大path.

    注意,当root == null,以上2个值都要置为Integer_MIN_VALUE; 因为没有节点可取的时候,是不存在solution的。以免干扰递归的计算

     1 /**
     2  * Definition for binary tree
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public class ReturnType {
    12         int maxSingle;
    13         int max;
    14         ReturnType (int maxSingle, int max) {
    15             this.max = max;
    16             this.maxSingle = maxSingle;
    17         }
    18     }
    19     
    20     public int maxPathSum(TreeNode root) {
    21         return dfs(root).max;        
    22     }
    23     
    24     public ReturnType dfs(TreeNode root) {
    25         ReturnType ret = new ReturnType(Integer.MIN_VALUE, Integer.MIN_VALUE);
    26         if (root == null) {
    27             return ret;
    28         }
    29         
    30         ReturnType left = dfs(root.left);
    31         ReturnType right = dfs(root.right);
    32         
    33         int cross = root.val;
    34         
    35         // if any of the path of left and right is below 0, don't add it.
    36         cross += Math.max(0, left.maxSingle);
    37         cross += Math.max(0, right.maxSingle);
    38         
    39         // 注意,这里不可以把Math.max(left.maxSingle, right.maxSingle) 与root.val加起来,
    40         // 会有可能越界!
    41         int maxSingle = Math.max(left.maxSingle, right.maxSingle);
    42         
    43         // may left.maxSingle and right.maxSingle are below 0
    44         maxSingle = Math.max(maxSingle, 0);
    45         maxSingle += root.val;
    46         
    47         ret.maxSingle = maxSingle;
    48         ret.max = Math.max(right.max, left.max);
    49         ret.max = Math.max(ret.max, cross);
    50         
    51         return ret;
    52     }
    53 }
    View Code

    GITHUB:

    https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/MaxPathSum.java

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  • 原文地址:https://www.cnblogs.com/yuzhangcmu/p/4172855.html
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