leetcode: Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

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本文介绍了一种寻找给定二叉树中最大路径和的算法实现。该路径可以始于并结束于树中的任意节点。通过递归方法计算每个子树的最大贡献，并更新全局最大路径和。

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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

Return 6.

this code is a little bit complicated, I will revise it soon

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#define INVALID_NUM 0xFFFFFFFF
 
class Solution {
    
    int maxPathSum2(TreeNode *root, int &maxVal)
    {
        if (root == NULL)
        {
            return INVALID_NUM;
        }
        
        int leftMax  = maxPathSum2(root->left, maxVal);
        int rightMax = maxPathSum2(root->right, maxVal);   
        int curMax;
        
        
        if (leftMax != INVALID_NUM && rightMax != INVALID_NUM)
        {
            if (leftMax <= 0 && rightMax <=0)
            {
                curMax = root->val;
            }
            else if (leftMax <= 0)
            {
                curMax = root->val+rightMax;
            }
            else if (rightMax <= 0)
            {
                curMax = root->val+leftMax;
            }
            else
            {
                if (root->val+leftMax >= root->val+rightMax)
                {
                    curMax = root->val+leftMax;
                }
                else
                {
                    curMax = root->val+rightMax;
                }
    
                if (root->val >= curMax)
                    curMax = root->val;
                    
                if (root->val+leftMax+rightMax > maxVal)
                {
                    maxVal = root->val+leftMax+rightMax;
                }
                return curMax;
            }
               
        }
        else if (leftMax != INVALID_NUM)
        {
            if (leftMax <= 0)
            {
                curMax = root->val;
            }
            else
            {
                curMax = root->val+leftMax;
            }
        }
        else if (rightMax != INVALID_NUM)
        {
            if (rightMax <= 0)
            {
                curMax = root->val;
            }
            else
            {
                curMax = root->val+rightMax;
            }
        }
        else
        {
            curMax = root->val;
        }
        
        if (curMax > maxVal)
            maxVal = curMax;

        return curMax;
    }
    
public:
    int maxPathSum(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() functionr
        int maxVal = -10000;
        maxPathSum2(root, maxVal);
        
        return maxVal;
    }
};