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.

              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;
        }
        
    }
}

Runtime: 254 ms

Path Sum