Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

转载最新推荐文章于 2024-03-25 21:39:45 发布 · 46 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zengzy/p/5056777.html

本文介绍了一种使用递归方法解决二叉树最大路径和问题的算法，详细解释了如何通过节点的值及其左右子树的路径和来计算整个二叉树的最大路径和，并提供了代码实现。

Binary Tree Maximum Path Sum

Total Accepted: 55016 Total Submissions: 246213 Difficulty: Hard

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 does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

(E) Path Sum (M) Sum Root to Leaf Numbers

设最大路径和为sum , 那么sum = max(sum,左子树根出发的最大路径和+根结点的值+右子树根出发的最大路径和);

根据这个式子写成递归式就可以了

有一点需要注意，如果从左右子树根出发的最大路径和为负数，那么将其值置0，因为根的值+从左右子树根出发的最大路径和一定会小于根的值

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode *root) {
        int maxPath = INT_MIN;
        maxPathWithNode(root,maxPath);
        return maxPath;
    }
    int maxPathWithNode(TreeNode* node,int& maxPath){
        if(!node) return 0;
        int l = max(0,maxPathWithNode(node->left,maxPath));
        int r = max(0,maxPathWithNode(node->right,maxPath));
        maxPath = max(maxPath,l+r+node->val);
       // cout<<"node->val="<<node->val<<" l="<<l<<" r="<<r<<" maxPath="<<maxPath<<endl;
        return max(l,r)+node->val;
    }
};
/*
[-1,-2,-3,4,-5,-6,-7,8,-9,-1,-1,-1,-1,-1,-1,-7,-8]
[-9,-9]
*/