LeetCode 112. Path Sum

Description:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Example:

For example:
Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

分析：

首先判断根节点是否为空，若为空，则返回false，表示肯定不能找到一条路径上的数之和等于sum；
若不空，则判断左右子树是否为空，若为空，则比较根节点的数是否等于sum，若为真，返回true；反之返回false；
若前面都不满足条件，则证明树的高度至少为2，递归左子树，将sum的值改成sum - root->val；递归右子树，将sum的值改成sum - root->val，即可求解。

代码如下：

#include <iostream>
using namespace std;
/**
 * Definition for a binary tree node.
 */
struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
	bool hasPathSum(TreeNode* root, int sum) {
		if (root == NULL) return false;
		if (root->left == NULL && root->right == NULL) return sum == root->val ? true : false;
		return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
	}
};
// 构造二叉树
int TreeNodeCreate(TreeNode* &tree) {
	int val;
	cin >> val;
	if (val < 0) // 小于0表示空节点
		tree = NULL;
	else {
		tree = new TreeNode(val); // 创建根节点
		tree->val = val;
		TreeNodeCreate(tree->left); // 创建左子树
		TreeNodeCreate(tree->right);// 创建右子树
	}
	return 0;
}
int main() {
	Solution s;
	TreeNode* tree;
	TreeNodeCreate(tree);
	int sum;
	cin >> sum;
	cout << s.hasPathSum(tree, sum) << endl;
	return 0;
}