[leetcode]Binary Tree Maximum Path Sum_左加根-优快云博客

本文链接：https://blog.youkuaiyun.com/menglinaoxiang/article/details/24192023

本文介绍了如何使用动态规划和后序遍历来解决二叉树中任意两个节点间的最大路径和问题。通过递归地遍历每个节点，并更新最大路径和，最终得到整个二叉树的最大路径和。

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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

题目：求任意两个结点间路径的最大和

解题思路：后续遍历二叉树与动态规划（一维数组求最大连续子数组和）结合运用。

    int innerMaxPathSum(TreeNode *root, int &maxValue){
        int v = root->val;
        int l, r;
        
        maxValue = max(v, maxValue);
        if(NULL == root->left && NULL == root->right){
            return v; //左右子数都为空 即叶结点的情况 直接返回其值 
        }
        if(NULL == root->right){ //只有左子树的情况
            l = innerMaxPathSum(root->left, maxValue);
            maxValue = max(l, maxValue);
            maxValue = max(v + l, maxValue);
            
            return max(v, l + v); //该种情况返回必须包含根结点的路径的最大值， 如果左加根比根小则丢弃根
        }
        if(NULL == root->left){
            r = innerMaxPathSum(root->right, maxValue);
            maxValue = max(r, maxValue);
            maxValue = max(v + r, maxValue); 
            
            return max(v, r + v);//同左
        }
        l = innerMaxPathSum(root->left, maxValue);
        r = innerMaxPathSum(root->right, maxValue);
        
        int oneMaxSum = max(l, r);
        maxValue = max(oneMaxSum, maxValue);
        
        int twoMaxSum = max(l + v, v + r);
        maxValue = max(twoMaxSum, maxValue);
        maxValue = max(l + v + r, maxValue);
        
        return max(twoMaxSum, v);  //同理要包含根结点， 此时有三种情况，要么是根，要么是根左，要么是根右  
    }
    
    int maxPathSum(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int maxValue = root->val;
        innerMaxPathSum(root, maxValue);
        
        return maxValue;
    }