LeetCode-Alg-897-Increasing Order Search Tree

LeetCode-Alg-897-Increasing Order Search Tree

Posted by Jae on May 19, 2019

1、题目

Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9   Note:

The number of nodes in the given tree will be between 1 and 100.
Each node will have a unique integer value from 0 to 1000.

2、思路

这道题意思是对一棵树进行重新排列,将最左边的节点作为树的根节点,每一个非叶子节点只有右孩子,没有左孩子,节点的值不需要按照大小排序,于是想到直接在树上做连接,将每个节点的左孩子全部断开置为null。

3、实现

3.1 栈实现中序非递归遍历

public class Solution {
    public TreeNode increasingBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        // 根节点
        TreeNode rootNode = null;
        // 保存前一个节点
        TreeNode preNode = null;
        boolean isRoot = true;

        while (root != null || !stack.isEmpty()){
            while (root != null){
                stack.push(root);
                if (root.left == null && isRoot){
                    rootNode = root;
                    preNode = root;
                }
                root = root.left;
            }

            root = stack.pop();
            if (isRoot){
                root = root.right;
                isRoot = false;
                continue;
            }
            // 左子树全部置为null
            root.left = null;
            preNode.right = root;
            preNode = root;
            root = root.right;
        }

        return rootNode;
    }
}

3.2 递归中序遍历

public class Solution {
    // 保存前一个节点
    private TreeNode current;

    public TreeNode increasingBST3(TreeNode root){
        TreeNode ans = new TreeNode(0);
        current = ans;
        inOrder(root);
        return ans.right;
    }
    public void inOrder(TreeNode root){
        if (root == null){
            return;
        }
        inOrder(root.left);
        root.left = null;
        current.right = root;
        current = root;
        inOrder(root.right);
    }
}