Leetcode 669 修剪二叉搜索树
作者:互联网
Leetcode 669 修剪二叉搜索树
数据结构定义:
给你二叉搜索树的根节点 root ,同时给定最小边界low 和最大边界 high。通过修剪二叉搜索树,使得所有节点的值在[low, high]中。修剪树不应该改变保留在树中的元素的相对结构(即,如果没有被移除,原有的父代子代关系都应当保留)。 可以证明,存在唯一的答案。
所以结果应当返回修剪好的二叉搜索树的新的根节点。注意,根节点可能会根据给定的边界发生改变。
示例 1:
1 1
/ \ ---> \
0 2 2
输入:root = [1,0,2], low = 1, high = 2
输出:[1,null,2]
示例 2:
3 3
/ \ ---> /
0 4 1
\
2
/
1
输入:root = [3,0,4,null,2,null,null,1], low = 1, high = 3
输出:[3,2,null,1]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
递归方式:
/*
* 思路: https://leetcode-cn.com/problems/trim-a-binary-search-tree/solution/669-xiu-jian-er-cha-sou-suo-shu-di-gui-die-dai-xia/
*/
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
if(root == null) return null;
if(root.val < low) return trimBST(root.right,low,high);
else if(root.val > high) return trimBST(root.left,low,high);
root.left = trimBST(root.left,low,high);
root.right = trimBST(root.right,low,high);
return root;
}
}
迭代方式:
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
if(root == null)
return null;
TreeNode node = root;
while(node.val < low || node.val > high){
if(node.val < low)
node = node.right;
else
node = node.left;
}
root = node;
while(root != null){
while(root.left != null && root.left.val < low){
root.left = root.left.right;
}
root = root.left;
}
root = node;
while(root != null){
while(root.right != null && root.right.val > high){
root.right = root.right.left;
}
root = root.right;
}
return node;
}
}
标签:right,TreeNode,669,val,二叉,low,root,Leetcode,left 来源: https://www.cnblogs.com/CodingXu-jie/p/14120588.html