leetcode 112 Path Sum

最新推荐文章于 2020-06-08 12:39:15 发布

walter1990

最新推荐文章于 2020-06-08 12:39:15 发布

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本文探讨了如何确定一棵二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的目标值。通过递归算法实现这一功能，并给出具体示例。

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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.

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.

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/**
 * 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 findSum(TreeNode *root, int remain) {
		if(root!=NULL && root->left==NULL && root->right==NULL) {
		     if(remain == root->val) 
			    return true;
			 else
			    return false;
		}

		if(root==NULL) return false;

		return findSum(root->left, remain-root->val) || findSum(root->right, remain-root->val);
	}
	bool hasPathSum(TreeNode *root, int sum) {
	    cout << sum << endl;
		return findSum(root, sum);
	}
};