Path Sum - LeetCode

最新推荐文章于 2019-05-04 10:13:03 发布

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本文提供了一个算法，用于判断给定的二叉树中是否存在从根节点到叶子节点的路径，其路径上的节点值之和等于指定的数值。通过递归遍历树的每个节点，计算从当前节点到叶子节点的路径和，从而判断是否存在满足条件的路径。

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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.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool AllNode(TreeNode *node,int subsum,int sum)
    {
	    subsum = subsum + node->val;
	    bool x;
	    if(node->left == NULL && node->right == NULL)
	    {
	    	if (subsum == sum)
		    {
			    return true;
		    }
		    else
			    return false;
		
	    }
	    if (node->left != NULL)
	    {
		    x = AllNode(node->left, subsum, sum);
		    if (x)
		    {
			    return true;
		    }
	    }
	    if (node->right != NULL)
	    {
		    return AllNode(node->right, subsum, sum);
	    }
    }

    bool hasPathSum(TreeNode *root, int sum) 
    {
        if (root == NULL)
	    {
		    return false;
	    }
	    int subsum = 0;
	    return AllNode(root, subsum, sum);
    }
};