# LeetCode – Path Sum Solution

### 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.

### 实现：

• 递归方式，DFS遍历所有路径找出符合条件的路径， sum值，每次调用时剪掉当前的值， 最后符合条件的路径中叶子节点的值与sum相同， 空间复杂度O(N)    时间复杂度 O(N)
``````
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null ){
return false;
}
if (root.left == null && root.right == null && root.val == sum){
return true;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
``````
• 迭代方式，使用栈实现先序遍历， 对于sum计算与递归模式不同，入栈前加上父节点的val， 这样如果找到符合要求的节点， 则节点val与sum值一致。     空间复杂度O(N)    时间复杂度 O(N)
``````
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null){
return false;
}
Stack<TreeNode> st = new Stack<TreeNode>();
st.push(root);

while (!st.empty()){
TreeNode node = st.pop();
if (node.left == null && node.right == null && node.val == sum){
return true;
}
if (node.left != null){
node.left.val += node.val;
st.push(node.left);
}
if (node.right != null){
node.right.val += node.val;
st.push(node.right);
}
}
return false;
}
}
``````