[LeetCode] 014: Binary Tree Maximum Path Sum

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1.02^365的成长裂变

最新推荐文章于 2024-03-25 21:39:45 发布

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分类专栏： LeetCode2

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本文介绍了一种求解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。通过递归遍历的方式，算法能够有效地找到从根节点到叶子节点的最大路径和。

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[Problem]

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.

[Solution]

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    // max path sum end with 'root'
    int maxPathSum(TreeNode *root, int &maxSum){
        if(root == NULL){
            return 0;
        }
        // max path sum end with root->left
        int left = maxPathSum(root->left, maxSum);
        
        // max path sum end with root->right
        int right = maxPathSum(root->right, maxSum);
        
        // update maxSum
        maxSum = max(maxSum, root->val);
        maxSum = max(maxSum, left + root->val);
        maxSum = max(maxSum, right + root->val);
        maxSum = max(maxSum, left + root->val + right);
        
        // max path sum end with root
        if(left < 0 && right < 0){
            return root->val;
        }
        else{
            return max(left, right) + root->val;
        }
    }
    
    // max path sum in subtree root
    int maxPathSum(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        // null tree
        if(root == NULL){
            return 0;
        }
        int res = root->val;
        maxPathSum(root, res);
        return res;
    }
};