Sum Root to Leaf Numbers 求所有二叉树根到叶子节点路径之和 @LeetCode

典型树形递归题，主要是要处理好根节点为null，叶子节点的情况

package Level3;

import Utility.TreeNode;

/**
 * Sum Root to Leaf Numbers 
 * 
 * Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.
 *
 */
public class S131 {

	public static void main(String[] args) {
		TreeNode root = new TreeNode(1);
		TreeNode n1 = new TreeNode(2);
		TreeNode n2 = new TreeNode(3);
		root.left = n1;
		root.right = n2;
		
		System.out.println(sumNumbers(root));
	}

	public static int sumNumbers(TreeNode root) {
		StringBuffer sb = new StringBuffer();
		return rec(root, sb);
	}
	
	public static int rec(TreeNode root, StringBuffer sb){
		if(root==null){
			return 0;
		}
		
		// 处理叶子节点情况
		if(root.left==null && root.right==null){
			// 临时添加入叶子节点然后再删除恢复现场
			int val = Integer.parseInt(sb.append(root.val).toString());
			sb.deleteCharAt(sb.length()-1);
			return val;
		}
		
		sb.append(root.val);
		int a = rec(root.left, sb);
		int b = rec(root.right, sb);
		sb.deleteCharAt(sb.length()-1);
		return a+b;
	}
}

Second try:

    public static int sumNumbers(TreeNode root) {
		return rec2(root, 0);
	}
	
	public static int rec2(TreeNode root, int n){
		if(root == null){
			return 0;
		}
		n = n*10 + root.val;
		if(root.left==null && root.right==null){
			return n;
		}
		return rec2(root.left, n) + rec2(root.right, n);
	}

/**
 * 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) {
        int[] sum = new int[1];
        recAdd(root, sum, 0);
        return sum[0];
    }
    
    public void recAdd(TreeNode root, int[] sum, int path) {
        if(root == null) {
            return;
        }
        
        if(root.left==null && root.right==null){
            path = path*10 + root.val;
            sum[0] += path;
            return;
        }
        
        path = path*10 + root.val;
        recAdd(root.left, sum, path);
        recAdd(root.right, sum, path);
    }
}