leetcode刷题笔记一百一十四题 二叉树展开为链表

leetcode刷题笔记一百一十四题 二叉树展开为链表

源地址：114. 二叉树展开为链表

问题描述：

给定一个二叉树，原地将它展开为一个单链表。

例如，给定二叉树

1

/
2 5
/ \
3 4 6
将其展开为：

1

2

3

4

5

6

/**
本题方法可以使用先序遍历方式，具体可分为递归和迭代
另一种方法可以使用先驱结点
*/
//递归方法 -- 先序遍历 将先序遍历结果放入res，然后修改左右结点
/**
 * Definition for a binary tree node.
 * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
 *   var value: Int = _value
 *   var left: TreeNode = _left
 *   var right: TreeNode = _right
 * }
 */
import scala.collection.mutable 
object Solution {
  def flatten(root: TreeNode): Unit = {
      var res = new mutable.ListBuffer[TreeNode]()
      preOrderTraversal(root)
      for (i <- 1 to res.size - 1){
          val prev = res(i-1)
          val cur = res(i)
          prev.left = null
          prev.right = cur
      }


      def preOrderTraversal(root: TreeNode): Unit = {
          if (root != null){
              res += root
              preOrderTraversal(root.left)
              preOrderTraversal(root.right)
          }
      }
  }
}

//与递归方法思想一致，使用stack实现迭代方法
/**
 * Definition for a binary tree node.
 * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
 *   var value: Int = _value
 *   var left: TreeNode = _left
 *   var right: TreeNode = _right
 * }
 */
import scala.collection.mutable 
object Solution {
  def flatten(root: TreeNode): Unit = {
        if (root == null) return
        val stack = new mutable.Stack[TreeNode]()
        stack.push(root)
        var prev: TreeNode = null
        while (stack.isEmpty == false){
            val cur = stack.pop
            if (prev != null){
                prev.left = null
                prev.right = cur
            } 
            
            val left = cur.left
            val right = cur.right

            if (right != null) stack.push(right)
            if (left  != null)  stack.push(left)
            prev = cur
        }
  }
}

//对pre cur next结点调整先序结点位置，本质还是按照先序遍历改变树的结构
/**
 * Definition for a binary tree node.
 * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
 *   var value: Int = _value
 *   var left: TreeNode = _left
 *   var right: TreeNode = _right
 * }
 */
import scala.collection.mutable 
object Solution {
  def flatten(root: TreeNode): Unit = {
    var cur = root
    while (cur != null){
        if (cur.left != null){
            var next = cur.left
            var predecessor = next
            while (predecessor.right != null) predecessor = predecessor.right
            predecessor.right = cur.right
            cur.left = null
            cur.right = next
        }
        cur = cur.right
    }
  }
}