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.

/**
* 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;
}
return inorder(root,0,sum);
}
public boolean inorder(TreeNode root ,int value , int sum){//value 是父节点传递下来的值，sum是要求达到的sum值
if(root.left == null && root.right == null){//叶子节点
if(value + root.val == sum){
return true;
}else{
return false;
}
}else{//非叶子节点
boolean left = false;
if(root.left != null){
left = inorder(root.left,root.val+value,sum);
}
boolean right = false;
if(root.right != null){
right = inorder(root.right,root.val+value,sum);
}
return left || right;
}

}
}

跟这一道题相比，还是比较简单的，思路很简单，一个DFS，找到叶子时也就找到了这条路径所有节点之和，然后判断即可。

Runtime: 254
ms