Path Sum

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.

Note: A leaf is a node with no children.

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.

Approach #1: C++.

/**
 * 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->val == sum && root->right == NULL && root->left == NULL) return true;
        return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
    }
};

　　

Approach #2: Java.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        if (root.val == sum && root.right == null && root.left == null) return true;
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
    }
}

　　

Approach #3: Python.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root == None:
            return False
        if root.val == sum and root.left == None and root.right == None:
            return True
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)