And the maximum path in the binary tree leetcode 124

Binary tree? And the maximum path!

Title:
Given a non-empty binary tree, and returns its maximum path.
In this problem, it is defined as a path from any node in the tree, to the sequence of any node. The path includes at least one node, and not necessarily through the root node.

Examples
Example 1:
Input: [1,2,3]

   1
  / \
 2   3

Output: 6

Example 2:
Input: [-10,9,20, null, null, 15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

Topic analysis

• Goal: to find a path in a binary tree node and its maximum ⇒

For a node
to go forward node ⇒ left / right node
path and the current node value + = residual path (to the left node / right starting node) and node values recursive ⇒

Problem-solving ideas

variable	effect
find()	To find the path and

process

1. The binary tree root in the queue s
2. ⇒ looking for a node to go to the left (to the left node as a starting point) and + right path to go (to the right node as a starting point) path and
3. Determining whether to update the path with the maximum
4. The left node is queued repeated 2,3 s

code show as below

int find(TreeNode*root)                                      //寻找 root 为起点的最大路径结点和
{
    if(!root) return 0;
    int left = 0, right = 0;
    if(!root->left && !root->right) return root->val;
    if(root -> left) left += find(root->left);
    if(root -> right) right += find(root->right);
    return max(max(root->val,root->val+left),root->val+right); 
}

class Solution {
public:
    int maxPathSum(TreeNode* root) {
        if(!root) return 0;
        int max0 = root->val;
        queue<TreeNode*> s{{root}};
        while(!s.empty())
        {
            TreeNode* temp = s.front();
            s.pop();
            int left = find(temp->left);        //向左走
            int right = find(temp->right);      //向右走
            max0 = max(max(max(temp->val,temp->val+left),max(temp->val+right,temp->val+left+right)),max0);        //判断是否更新最大值
            if(temp->left) s.push(temp->left);
            if(temp->right) s.push(temp->right);
        }
        return max0;
    }
};