# LeetCode-Alg-112-Path-Sum

## LeetCode-Alg-112-Path-Sum

Posted by Jae on February 27, 2019

### 1、题目

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.


### 2、思路

题目中让我们判断从根节点到叶子节点是否存在一条路径的和等于sum，我们可以认为一棵树是又一个树根和若干子树构成，我们用sum-root.val的数值，



### 3、递归实现

public boolean hasPathSum(TreeNode root, int sum) {
if (root == null)
{
return false;
}
// 如果是叶子节点直接返回
if (root.left==null && root.right==null)
{
return root.val == sum;
}

boolean res = true;
// 判断左子树是否满足
if (root.left != null)
{
// 如果是左叶子节点
if (root.left.left == null && root.left.right == null )
{
if ((sum - root.val) == root.left.val)
{
return true;
}
res = false;
}
// 非叶子节点
else{
res = hasPathSum(root.left, sum-root.val);
}

if (res)
{
return res;
}
}
// 如果左子树不满足再判断右子树是否满足
if (root.right != null)
{
// 如果是右叶子节点
if (root.right.left == null && root.right.right == null)
{
if ((sum - root.val) == root.right.val)
{
return true;
}
res = false;
}else
{
res =  hasPathSum(root.right, sum-root.val);
}

if (res)
{
return res;
}
}
return res;
}