LeetCode-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 · 74 阅读

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原文链接：http://www.cnblogs.com/superzrx/p/3350778.html

本文介绍了一种寻找二叉树中最大路径和的算法实现。通过递归方式遍历树的每个节点，计算从该节点开始的最大路径和，并找出整棵树中的最大值。文章提供了详细的C++代码示例。

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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(int&height,TreeNode *root){
        int l1,l2,h1,h2,l3;
        if(root->left!=NULL&&root->right!=NULL){
            l1=maxPathSum(h1,root->left);
            l2=maxPathSum(h2,root->right);
            l3=h1+h2+root->val;
            height=max(max(h1,h2)+root->val,root->val);
            return max(max(max(l1,l2),l3),height);
        }
        else if(root->right==NULL&&root->left==NULL){
            height=root->val;
            return root->val;
        }
        else{
            if(root->left!=NULL){
                l1=maxPathSum(h1,root->left);
                height=max(h1+root->val,root->val);
                return max(max(l1,height),root->val);
            }
            else{
                l2=maxPathSum(h2,root->right);
                height=max(h2+root->val,root->val);
                return max(max(l2,height),root->val);
            }
        }
    }
    int maxPathSum(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int height;
        if(root==NULL)return 0;
        return maxPathSum(height,root);
    }
};