LeetCode 124. Binary Tree Maximum Path Sum

原题链接在这里：https://leetcode.com/problems/binary-tree-maximum-path-sum/

题目：

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

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

题解：

dfs函数是求当前root往下的最大路径. DFS states needs current tree node.

先算左边路径，返回值直接和0比较, 若不是正数就返回0, 然后同样方法算右边路径.

更新res[0], 看看当前root的val同时加上左边和右边有没有比现有的res[0]包含值大.

dfs 返回当前最深路径看当前root的val加上左边或者右边, 哪个大, dfs函数返回大的那个.

Time Complexity: O(n).

Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int maxPathSum(TreeNode root) {
        if(root == null){
            return 0;
        }
        int [] res = {Integer.MIN_VALUE};
        dfs(root, res);
        return res[0];
    }
    
    private int dfs(TreeNode root, int [] res){
        if(root == null){
            return 0;
        }
        //自底向上的death-first
        int leftMax = Math.max(0, dfs(root.left, res));
        int rightMax = Math.max(0, dfs(root.right, res));
        res[0] = Math.max(res[0], leftMax + rightMax + root.val);
        
        return Math.max(leftMax, rightMax) + root.val;
    }
}

AC C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int res = INT_MIN;
        dfs(root, res);
        return res;
    }

private: 
    int dfs(TreeNode* root, int& res){
        if(!root){
            return 0;
        }

        int left = max(0, dfs(root->left, res));
        int right = max(0, dfs(root->right, res));
        res = max(res, left + right + root->val);
        return max(left, right) + root->val;
    }
};

类似Diameter of Binary Tree, Longest Univalue Path, Path Sum, Sum Root to Leaf Numbers.