【LeetCodeOJ】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.

java code : 448 ms

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == null)
            return false;
        return hasSum(root, 0, sum);
    }
    public boolean hasSum(TreeNode root, int tmp, int sum)
    {
        if(root.left == null && root.right == null)
        {
            return (tmp + root.val) == sum;
        }
        boolean isleft = false, isright = false;
        if(root.left != null)
        {
            isleft = hasSum(root.left, tmp + root.val, sum);
        }
        if(root.right != null)
        {
            isright = hasSum(root.right, tmp + root.val, sum);
        }
        return (isleft || isright);
    }
}

cpp code : 64 ms

/**
 * 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 == NULL)
            return false;
        return hasSum(root, 0, sum);
    }
    bool hasSum(TreeNode *root, int tmp, int sum)
    {
        if(root->left == NULL && root->right == NULL)
        {
            return (tmp + root->val) == sum;
        }
        bool isleft = false, isright = false;
        if(root->left != NULL)
        {
            isleft = hasSum(root->left, tmp + root->val, sum);
        }
        if(root->right != NULL)
        {
            isright = hasSum(root->right, tmp + root->val, sum);
        }
        return isleft || isright;
    }
};


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