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E

tags: Tree, DFS

给一个inputSum, 然后dfs, 找到是否有一条path, 得出的path sum 跟 inputSum 一样.

#### DFS

- 确定好结尾条件: `is leaf` && `val == sum`.

- 每一层减掉node.val, 然后dfs.

- 写一写: `root == null => false` 对parent nodes的影响. 这里发现没影响, 所以可以简化成用1个functionDFS.

```

/*

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.

*/

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

/*

Thoughts:

DFS on the function itself, keep subtracting the root.val.

when root == null && sum == 0, return true;

*/

class Solution {

public boolean hasPathSum(TreeNode root, int sum) {

if (root == null) return false;

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

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

}

}

```