2014
11-19

# LeetCode-Path Sum[二叉树]

### 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.

// LeetCode, Path Sum
// 时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum) {
if (root == nullptr) return false;

if (root->left == nullptr && root->right == nullptr) // leaf
return sum == root->val;

return hasPathSum(root->left, sum - root->val)
|| hasPathSum(root->right, sum - root->val);
}
};


Java代码:

/**
* 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) {
if(root == null)
return false;

if(root.left == null && root.right == null)
return (root.val == sum);

return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
}
}

Path Sum II

1. 是穷举，但是代码有优化（v数组），并不是2^n。测试数据应该没问题，之前有超时的代码。