112. 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 a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public boolean hasPathSum(TreeNode root, int sum) {
         boolean flag=false;
 
         if(root==null)
         return false;
 
         if(root.left==null&&root.right==null&&root.val==sum)
         {
         flag=true;
         return flag;
         }
 
         if(root.left!=null||root.right!=null)
         {
             if(root.left!=null&&flag==false)
             {
              flag=hasPathSum(root.left,sum-root.val);
              if(flag==true)
              return flag;
             }
 
             if(root.right!=null&&flag==false)
             {
              flag=hasPathSum(root.right,sum-root.val);
              if(flag==true)
              return flag;
             }
 
         }
         return flag;
     }
 }

            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.

代码如下：