leetcode_129_Sum Root to Leaf Numbers

来源：互联网发布：ubuntu 安装luajit 编辑：程序博客网时间：2024/05/29 18:34

描述：

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.

思路：

和pathsum系列是一个思路，后序遍历至某个节点，将root-to-leaf的所有结点的值表示成一个数和sum相加，直至遍历完所有的leaf结点，返回sum。

代码：

public int sumNumbers(TreeNode root) {if (root == null)return 0;int temp=0,sum=0;LinkedList<TreeNode>linkedList=new LinkedList<TreeNode>();Stack<TreeNode> st1 = new Stack<TreeNode>();linkedList.addLast(root);TreeNode top = null;while (!linkedList.isEmpty()) {top=linkedList.getLast();while(top.left!=null)//达到页结点{linkedList.addLast(top.left);top=top.left;}while(!linkedList.isEmpty()){top=linkedList.getLast();if(top.right==null)//无右子树时，访问结点{if(top.left==null){temp=0;for(int i=0;i<linkedList.size();i++)temp=temp*10+linkedList.get(i).val;sum+=temp;}linkedList.removeLast();}else {if(st1.empty()||(!st1.empty()&&st1.peek()!=top))//有右子树且第一次出现，将右子树入栈，并将结点在st1中标记一下{st1.push(top);linkedList.addLast(top.right);break;//右子树入栈了，然后从右子树开始}else//有右子树，但是第二次出现了，访问该结点{st1.pop();linkedList.removeLast();}}}}return sum;    }

结果：

0 0