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  • LeetCode-Binary Tree Maximum Path Sum

    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);
        }
    };
    View Code
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  • 原文地址:https://www.cnblogs.com/superzrx/p/3350778.html
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