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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
.
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
Binary Tree Maximum Path Sum
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