LeetCode125 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.

解题思想：二叉树遍历，判断如果是叶子节点，就把路径的节点转换为整数。这里面我用了二个栈

stack用于返回上个节点

stack1用于记录每次根节点到叶子的所有节点，stack1中节点和stack一样添加节点，但是弹出的时机不一样。要注重分析。

<span style="color:#333333;"> public int sumNumbers(TreeNode root) {
        if(root==null)return 0;
        Stack<TreeNode> stack=new Stack<TreeNode>();
        Stack<TreeNode> stack1=new Stack<TreeNode>();
        StringBuilder builder=new StringBuilder();
        int sum=0;
        TreeNode current=root;
        while(current!=null||!stack.isEmpty())
        {
            if(current!=null)
            {
                stack.push(current);
                stack1.push(current);
                current=current.left;
            }else{
                current=stack.pop();
                if(current.right==null&¤t.left==null)
                {
                    for(int i=0;i<stack1.size();i++)
                    {
                    builder.append(stack1.get(i).val);
                    }
                    sum+=Integer.parseInt(builder.toString());
                    builder.delete(0, builder.length());
                    </span><span style="color:#ff0000;">stack1.pop();//叶子节点计算完弹出</span><span style="color:#333333;">
                }else{
                    </span><span style="color:#ff0000;">while(!stack1.isEmpty())
                    {
                     if(stack1.peek()==current)break;//当current存在右节点时stack中的current的节点已经弹出
                     stack1.pop();                  //但是stack1中的current节点不能弹出，我们要保持路径上的完整节点
                    }</span><span style="color:#333333;">
                }
                current=current.right;
            }
        }
        return sum;
    }</span>