标签：convert == roo lse int null ret amp sem
题目描述：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。
实现语言：Java
/**
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
*/
public class Solution {
public TreeNode Convert(TreeNode root) {
if(root==null||root.left==null&&root.right==null){
return root;
}
TreeNode left=Convert(root.left);
TreeNode node=left;
while(node!=null&&node.right!=null){
node=node.right;
}
if(node!=null){
node.right=root;
root.left=node;
}
TreeNode right=Convert(root.right);
if(right!=null){
root.right=right;
right.left=root;
}
return left!=null?left:root;
}
}
实现语言：Java
/**
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
*/
import java.util.Stack;
public class Solution {
public TreeNode Convert(TreeNode root) {
if(root==null||root.left==null&&root.right==null){
return root;
}
Stack<TreeNode> stk=new Stack<TreeNode>();
TreeNode node=null;
while(root!=null||!stk.isEmpty()){
if(root!=null){
stk.push(root);
root=root.right;
}else{
root=stk.pop();
if(node==null){
node=root;
}else{
node.left=root;
root.right=node;
node=root;
}
root=root.left;
}
}
return node;
}
}
标签：convert == roo lse int null ret amp sem
原文地址：https://www.cnblogs.com/xidian2014/p/10197904.html
