[LeetCode]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.

 

题挺水的,不过刚开始一直wa了几次,发觉基础真的好重要,我把leaf这个概念弄混了.我刚开始只是判断有一条路某时刻满足条件就判断true了，结果就wa了好几次.

leaf的定义是左右子节点都为空！！！

 

code

 

public class Solution {

    public boolean hasPathSum( TreeNode root, int sum ) {
        if( root != null ) {
            sum -= root.val;
            if( sum == 0 && root.left == null && root.right == null )
                return true;
            else
                return hasPathSum( root.left, sum ) || hasPathSum( root.right, sum );
        } else
            return false;
    }

    public static void main( String[] args ) {
        Solution s = new Solution();
        TreeNode root = new TreeNode( 5 );
        root.left = new TreeNode( 4 );
        root.left.left = new TreeNode( 11 );
        root.left.left.left = new TreeNode( 7 );
        root.left.left.right = new TreeNode( 2 );
        root.right = new TreeNode( 8 );
        root.right.left = new TreeNode( 13 );
        root.right.right = new TreeNode( 4 );
        root.right.right.right = new TreeNode( 1 );

        // root.left = new TreeNode( 1 );

        System.out.println( s.hasPathSum( root, 9 ) );
    }

}

class TreeNode {

    int val;

    TreeNode left;

    TreeNode right;

    TreeNode( int x ) {
        val = x;
    }
}