LeetCode - Path Sum

Path Sum

2014.1.8 04:59

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.

Solution:

　　This problem has no constraint on the value of the tree nodes, neither being sorted nor limited to a certain range. Thus we can only perform a thorough recursive search. Note that it's root-to-leaf, so don't stop halfway on non-leaf nodes.

　　Time and space complexities are both O(n), where n is the number of nodes in the tree. Space complexity comes from local parameters passed in function calls.

Accepted code:

/**
 * 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 hasPathSum(TreeNode *root, int sum) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == nullptr){
            return false;
        }

        if(root->left == nullptr){
            if(root->right == nullptr){
                return root->val == sum;
            }else{
                return hasPathSum(root->right, sum - root->val);
            }
        }else{
            if(root->right == nullptr){
                return hasPathSum(root->left, sum - root->val);
            }else{
                return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
            }
        }
    }
};