LintCode 475: Binary Tree Maximum Path Sum II

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本文探讨了在二叉树中寻找从根节点到任意节点的最大路径和问题，通过深度优先搜索（DFS）策略实现。示例展示了输入为{1,2,3}

Binary Tree Maximum Path Sum II
中文English
Given a binary tree, find the maximum path sum from root.

The path may end at any node in the tree and contain at least one node in it.

Example
Example 1:

Input: {1,2,3}
Output: 4
Explanation:
1
/
2 3
1+3=4
Example 2:

Input: {1,-1,-1}
Output: 1
Explanation:
1
/
-1 -1

解法1：DFS。
注意：
1）当左右最大分枝返回负数时，此时用root->val即可。

代码如下：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: the root of binary tree.
     * @return: An integer
     */
    int maxPathSum2(TreeNode * root) {
        if (!root) return 0;
        int result = 0;        
        result = max(maxPathSum2(root->left), maxPathSum2(root->right));
        if (result < 0) return root->val;
        else return result + root->val;
    }
};