Flatten Binary Tree to Linked List

最新推荐文章于 2022-09-06 16:04:14 发布

转载最新推荐文章于 2022-09-06 16:04:14 发布 · 321 阅读

Leetcode 同时被 2 个专栏收录

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本文介绍了一种使用递归方法实现二叉树的前序展平算法，并提供了一个混合迭代与递归的方法。此外，还展示了如何在二叉搜索树中填充每个节点到其后继节点的引用及查找第X个节点。

注意要传previous node，而java中不能传reference

法一：prev为static variable 法二：用个array wrap它...

1. Preorder flatten(Leetcode):

public class Solution { 
    TreeNode prev = null;
    public void flatten(TreeNode root) { 
        if (root == null) {
            return;
        }
        TreeNode rightnode = root.right;
        if (prev != null) {
            prev.left = null;
            prev.right = root;
        }
        
        prev = root;
        flatten(root.left);
        
       // prev = root.left;
        flatten(rightnode);
    } 
}

之前写的混有iteration的recursion

public class Solution { 
    public void flatten(TreeNode root) { 
       // use recursion
        if(root == null) return;
        
        TreeNode left = root.left;
        TreeNode right = root.right;
        
        flatten(root.left);
        flatten(root.right);

        root.right = left;
        root.left = null;
        
        TreeNode temp = root;
        while(temp.right!=null){
            temp = temp.right;
        }
        temp.right = right;
    } 
}

2. populating each node to its successor in BST

public void addSuccessor(Node node, Node[] prev) {
  if (node == null)
    return;
  addSuccessor(node.left, prev);
  if (prev[0] != null)
    prev[0].next = node;
  prev[0] = node;
  addSuccessor(node.right, prev);
}

3. find the xth node in BST

public TreeNode getNthNode(TreeNode node, IntWrapper i, int x) {
  if (node == null)
    return null;
  TreeNode foundAtLeft = getNthNode(node.left, i, x);
  i.val++;
  if (i.val == x)
     return node;
  if (foundAtLeft != null)
    return foundAtLeft;
    return getNthNode(node.right, i, x);
}