Leetcode_path-sum (updated c++ and python version)

地址：http://oj.leetcode.com/problems/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.

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.

思路：深度优先遍历dfs。注意根为NULL和只有一个根节点情况。要把数据结构里的基础知识变通一下。

参考代码：

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

//SECOND TRIAL
class Solution {
private:
    bool dfs(TreeNode* root, int left)
    {
        if(left-root->val == 0 && !root->left && !root->right)
            return true;
        else
        {
            if(root->left && dfs(root->left, left-root->val))
                return true;
            if(root->right && dfs(root->right, left-root->val))
                return true;
            return false;
        }
    }
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if(!root)
            return false;
        return dfs(root, sum);
    }
};

python :

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

class Solution:
    def dfs(self, root, left):
        if root.val==left and not root.left and not root.right:
            return True
        if root.left and self.dfs(root.left, left-root.val):
            return True
        if root.right and self.dfs(root.right, left-root.val):
            return True
        return False
    # @param root, a tree node
    # @param sum, an integer
    # @return a boolean
    def hasPathSum(self, root, sum):
        if not root:
            return False
        return self.dfs(root, sum)