【LeetCode】 Sum Root to Leaf Numbers

标签：leetcode   递归   

第一次写的解法，。对于只有一个儿子的节点重复了两次 。。结果就弄复杂了。。我也觉得不应该能有这么多的重复嘛

<span style="font-size:18px;">/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */ 
package javaTrain;

public class Train9 {
	public int sum = 0;
    public int sumNumbers(TreeNode root) {
    	if(root == null) return sum;
    	sum = dfs(root.left,root.val)+dfs(root.right,root.val);
    	return sum;
    } 
    private int dfs(TreeNode root,int num){
    	if(root == null) return num;
    	return dfs(root.left,num*10+root.val) + dfs(root.right,num*10+root.val);
    }
}
</span>

调整后的正解，唉~边界要想清楚呀。。

<span style="font-size:18px;">/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumNumbers(TreeNode root) {
      return helper(root,0);
    }
private int helper(TreeNode root, int sum)
{
    if(root == null)
        return 0;
    if(root.left==null && root.right==null)
        return sum*10+root.val;
    return helper(root.left,sum*10+root.val)+helper(root.right,sum*10+root.val);
}
}</span>

【LeetCode】 Sum Root to Leaf Numbers

标签：leetcode   递归   

原文地址：http://blog.csdn.net/ymzmdx/article/details/44347511