Sum Root to Leaf Numbers

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.

这道题，我是在先序遍历的基础上改的，添加了一个类似堆栈的list，如果是非叶子节点将其“入栈”，如果是叶子节点node“弹栈”直到node的父亲节点出现，扫描“栈底”到“栈顶”所有元素组成的数字，计算sum。直到前序遍历结束

import java.util.ArrayList;
import java.util.List;




   class TreeNode {
      int val;
      TreeNode left;
      TreeNode right;
      TreeNode(int x) { val = x; }
  }

public class Solution {

    List<TreeNode> list = new ArrayList<TreeNode>();
    
    int sum = 0;
    
    public int sumNumbers(TreeNode root) {
        if(null == root)                                //空树返回0
            return 0;
        else if(null == root.left && null == root.right){//只有根节点返回根节点的值
            return root.val;
        }else{
            preOrder(root);                                //至少有一个子树
//            list.add(root);                            //根节点入栈
        }
        
        return sum;
    }
    
    /**
     * 前序遍历树
     * @param root
     */
    public void preOrder(TreeNode root){
        if(root == null)
            return;
        if(null == root.left && root.right == null){//叶子节点
            while(list.size() != 0 && root != list.get(list.size() - 1).left && root != list.get(list.size() - 1).right){
                list.remove(list.size() - 1);
            }
            int temp = 0;
            for(int i = 0; i < list.size(); i++){    //计算一个数
                temp = temp * 10 + list.get(i).val;
            }
            temp = temp * 10 + root.val;            //叶子节点加上去
            //System.out.println("temp = " + temp);
            sum += temp;                            //计算和
            if(root == list.get(list.size() - 1).right){    //如果叶子节点是右子树，出栈
                list.remove(list.size() - 1);
            }
        }
        else{    //非叶子节点
            while(list.size() != 0 && root != list.get(list.size() - 1).left && root != list.get(list.size() - 1).right){
                list.remove(list.size() - 1);
            }                        //root应该与栈顶元素相同
            list.add(root);
            preOrder(root.left);
            preOrder(root.right);
        }
    }
}

题目提示的是用DFS，一会儿百度一下~

看了一下discuss里面的，这个写的挺漂亮的

public class Solution {
    public int sumNumbers(TreeNode root) {
        return preOrder(root, 0);
    }    
    public int preOrder(TreeNode root, int val){
        if(root == null)
            return 0;
        val = val * 10 + root.val;
        if(null == root.left && null == root.right)
            return val;
        else
            return preOrder(root.left, val) + preOrder(root.right, val);
    }
}