LeetCode之Two Sum IV Input is a BST（Kotlin）-白红宇

问题： Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

方法：与TwoSum的解法差不多，不过需要在其中加入树的遍历，深度优先遍历和广度优先遍历都可以。把遍历到的值存入到map中，如果在后面能够找到与map中的值相加等于target的值就返回true，否则遍历结束后返回false。

具体实现：

class TwoSumIV {    private val map = mutableMapOf
    
     ()    private var target = 0    class TreeNode(var `val`: Int = 0) {        var left: TreeNode? = null        var right: TreeNode? = null    }    /**     * Definition for a binary tree node.     * class TreeNode(var `val`: Int = 0) {     *     var left: TreeNode? = null     *     var right: TreeNode? = null     * }     */    fun findTarget(root: TreeNode?, k: Int): Boolean {        target = k        return find(root)    }    private fun find(root: TreeNode?): Boolean {        if (root == null) {            return false        }        if (map[root.`val`] != null) {            return true        }        map.put(target - root.`val`, 1)        return find(root.left) || find(root.right)    }}/** *      5 *     / \ *    3   6 *   / \   \ *  2   4   7 */fun main(args: Array
     
      ) {    val root = TwoSumIV.TreeNode(5)    root.left = TwoSumIV.TreeNode(3)    root.right = TwoSumIV.TreeNode(6)    (root.left)?.left = TwoSumIV.TreeNode(2)    (root.left)?.right = TwoSumIV.TreeNode(4)    (root.right)?.right = TwoSumIV.TreeNode(7)    val twoSumIV = TwoSumIV()    val result = twoSumIV.findTarget(root, 15)    println("result $result")}复制代码

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