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 {
    boolean has;
    public boolean hasPathSum(TreeNode root, int sum) {
        // Start typing your Java solution below
        // DO NOT write main() function
        has = false;
        path(root, sum);
        return has;
    }
    public void path(TreeNode r, int s){
        if(r == null) return;
        if(r.left == null && r.right == null){
            if(r.val == s)has = true;
                return;
        }
        else{
            path(r.right, s - r.val);
            path(r.left, s - r.val);
        }
    }
}

 第二遍：

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == null) return false;
        sum -= root.val;
        if(root.left == null && root.right == null){
            if(sum == 0) return true;
            else return false;
        }
        return hasPathSum(root.left , sum) || hasPathSum(root.right, sum);
    }
}

if(root == null) return false; 很重要！