LeetCode 124:Binary Tree Maximum Path Sum

Datetime:2016-08-23 01:23:14          Topic: LeetCode           Share

Given a 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 does not need to go through the root.

For example:Given the below binary tree,

1
      / \
     2   3

Return 6 .

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//找出二叉树任意一点到另一点的路径,使得和最大.
//解题思路:后序遍历,先计算左右子树的值l和r,若l<0或r<0,则不用加上
class Solution {
public:
	int maxSum;
	int maxPathSum(TreeNode* root) {
		if (root == NULL) return 0;
		maxSum =root->val;
		dfs(root);
		return maxSum;
	}

	int dfs(TreeNode* root){
		if (root == NULL)
			return 0;
		int l = dfs(root->left);
		int r = dfs(root->right);
		int sum = root->val;
		if (l > 0) sum = sum + l;
		if (r > 0) sum = sum + r;
		maxSum = max(maxSum, sum);
		return max(root->val, max(l + root->val, r + root->val));
	}
};
#include<iostream>  
#include<new>  
#include<vector>  
#include<algorithm>
using namespace std;

//Definition for binary tree  
struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
//找出二叉树任意一点到另一点的路径,使得和最大.
//解题思路:后序遍历,先计算左右子树的值l和r,若l<0或r<0,则不用加上
class Solution {
public:
	int maxSum;
	int maxPathSum(TreeNode* root) {
		if (root == NULL) return 0;
		maxSum =root->val;
		dfs(root);
		return maxSum;
	}

	int dfs(TreeNode* root){
		if (root == NULL)
			return 0;
		int l = dfs(root->left);
		int r = dfs(root->right);
		int sum = root->val;
		if (l > 0) sum = sum + l;
		if (r > 0) sum = sum + r;
		maxSum = max(maxSum, sum);
		return max(root->val, max(l + root->val, r + root->val));
	}
};
	void createTree(TreeNode *&root)
	{
		int i;
		cin >> i;
		if (i != 0)
		{
			root = new TreeNode(i);
			if (root == NULL)
				return;
			createTree(root->left);
			createTree(root->right);
		}
	}

int main()
{
	Solution s;
	TreeNode *root;
	createTree(root);
	int sum = s.maxPathSum(root);
	cout << sum << endl;
	system("pause");
	return 0;
}




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