（Java）leetcode-897 Increasing Order Search Tree

题目描述

Given a binary search 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.

思路1

中序遍历二叉树，同时新建一颗二叉树，随着遍历的进行，不断往右孩子中填值。

代码

class Solution {
	private TreeNode newTree = new TreeNode();
	private TreeNode index = newTree;
    public TreeNode increasingBST(TreeNode root) {   	
        if (root == null) return null;
        increasingBST(root.left);
        newTree.right= new TreeNode(root.val);
        newTree = newTree.right;
        increasingBST(root.right);
        return index.right;
    }
}

思路2

不用新建树，直接修改指针。

代码

class Solution {
	TreeNode prev=null, head=null;
    public TreeNode increasingBST(TreeNode root) {
        if(root==null) return null;   
        increasingBST(root.left);  
        if(prev!=null) { 
        	root.left=null; // we no  longer needs the left  side of the node, so set it to null
        	prev.right=root; 
        }
        if(head==null) head=root; // record the most left node as it will be our root
        prev=root; //keep track of the prev node
        increasingBST(root.right); 
        return head;
    }
}