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任意节点，通过递归遍历左子树和右子树，计算每个节点的最大贡献值和当前节点路径的最大值，最终返回整棵树的最大路径和。

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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.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode *root) {
        max = INT_MIN;
        buildMaxMap(root);
        return max;
        
        
    }
    int buildMaxMap(TreeNode* node)
    {
        if(node == NULL)
            return 0;
        int maxL = buildMaxMap(node->left);
        int maxR = buildMaxMap(node->right);
        int maxNode = std::max(std::max(maxL, maxR) + node->val, node->val);
        int curMax = std::max(std::max(maxL + maxR + node->val, node->val), maxNode);
        if(curMax > max)
            max = curMax;
        return maxNode;
        
    }
private:
int max;

    
};

Round 2:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode *root) {
        result = INT_MIN;
        helper(root);
        return result;
    }
private:
    int helper(TreeNode *root)
    {
        if(root == NULL)
            return 0;
        int left = helper(root->left);
        int right = helper(root->right);
        int curMax = 0, localMax = 0;
        curMax = std::max(std::max(left, right) + root->val, root->val);
        localMax = std::max(curMax, left + right + root->val);
        if(localMax > result)
            result = localMax;
        return curMax;
        
    }
    int result;
    
};