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
.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int max(int a, int b) 13 { 14 return a > b? a : b; 15 } 16 17 int maxPathSum(TreeNode *root) { 18 if (!root) 19 return 0; 20 int val,ll,rr; 21 ll = rr = 0; 22 val = -65535; 23 if (root->left) 24 { 25 val = max(val,maxPathSum(root->left)); 26 ll = max(0,root->left->val); 27 } 28 if (root->right) 29 { 30 val = max(val,maxPathSum(root->right)); 31 rr = max(0,root->right->val); 32 } 33 val = max(val,root->val + ll + rr); 34 root->val += max(ll,rr); 35 return val; 36 } 37 };
先计算左子节点到叶子节点最大值,在计算右子节点到叶子节点最大值,最后计算根节点最大值,val代表所求最大值
要更新当前节点到叶节点的最大路径值,考虑负数情况