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.

 

思路：使用递归。如果当前节点是叶子节点，则看节点的值和sum是否相等。如果当前节点不是节点，则去找左右子树中是否存在这么条路径。

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null) return false;
        if(root.left==null&&root.right==null) return root.val==sum;
        return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);//左右子树只要有一个存在就行。。
    }
    
}