# 112 Path Sum

## 112. Path Sum

## 1. Question

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.

## 2. Implementation

(1) Recursion

```java
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;

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

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

(2) Iteration

```java
public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }

        Stack<TreeNode> nodes = new Stack<>();
        Stack<Integer> pathSums = new Stack<>();
        nodes.push(root);
        pathSums.push(root.val);

        while (!nodes.isEmpty()) {
            TreeNode curNode = nodes.pop();
            int curSum = pathSums.pop();

            if (curNode.left == null && curNode.right == null && curSum == sum) {
                return true;
            }

            if (curNode.left != null) {
                nodes.push(curNode.left);
                pathSums.push(curSum + curNode.left.val);
            }

            if (curNode.right != null) {
                nodes.push(curNode.right);
                pathSums.push(curSum + curNode.right.val);
            }
        }
        return false;
    }
```

## 3. Time & Space Complexity

Recursion: 时间复杂度: O(n), 空间复杂度:O(n)

Iteration: 时间复杂度: O(n), 空间复杂度: O(n)
