LeetCode之Construct String from B

问题：
You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.

The null node needs to be represented by empty parenthesis pair "()". And you need to omit all the empty parenthesis pairs that don't affect the one-to-one mapping relationship between the string and the original binary tree.

方法：
通过递归实现先根遍历，也可以用栈结构实现。然后根据题目的要求当左节点存在但右节点为空时省略右节点的（），左右节点都为空时省略所有括号，其他情况均保留括号。

具体实现：

class ConstructStringFromBinaryTree {
    class TreeNode(var `val`: Int = 0) {
        var left: TreeNode? = null
        var right: TreeNode? = null
    }

    fun tree2str(t: TreeNode?): String {
        if (t == null) {
            return ""
        }
        val sb = StringBuilder()
        sb.append(t.`val`)
        if (t.left != null && t.right == null) {
            sb.append("(${tree2str(t.left)})")
        } else if (t.right != null) {
            sb.append("(${tree2str(t.left)})")
            sb.append("(${tree2str(t.right)})")
        }
        return sb.toString()
    }
}

fun main(args: Array<String>) {
    val root = ConstructStringFromBinaryTree.TreeNode(5)
    root.left = ConstructStringFromBinaryTree.TreeNode(3)
    root.right = ConstructStringFromBinaryTree.TreeNode(6)
    (root.left)?.left = ConstructStringFromBinaryTree.TreeNode(2)
    (root.left)?.right = ConstructStringFromBinaryTree.TreeNode(4)
    (root.right)?.right = ConstructStringFromBinaryTree.TreeNode(7)
    val constructStringFromBinaryTree = ConstructStringFromBinaryTree()
    val result = constructStringFromBinaryTree.tree2str(root)
    println("result $result")
}

有问题随时沟通

具体代码实现可以参考Github