leetcode Question 66: Path Sum I

Path Sum I:

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.

Analysis:

This is a basic practice for depth first search algorithm.


The way of thinking this problem:

• This question requires only true/false return, so the final result is the "or" operation of all the possible root-to-leaf path.
• Parameters used in recursion should only be the sum value and the current node.
• Condition:  if the sum == 0 and the node is leaf node return true,
•                         if the sum != 0 and the node is NULL, return false
•                         else return go root.left, sum-val || go root.right, sum-val
• How to check if the current node is a leaf node?  Both its left and right children all empty.

Code(C++):

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root==NULL) {return false;}
        if ( (root->left==NULL) && (root->right==NULL) && (sum-root->val==0) ) {return true;}
        return (hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val));
    }
};

Code(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 not root:
            return False
        else:
            sum = sum - root.val
            if not root.left and not root.right:
                if sum==0:
                    return True
            return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)