死磕算法系列文章
“Leetcode : https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof
“GitHub : https://github.com/nateshao/leetcode/blob/main/algo-notes/src/main/java/com/nateshao/sword_offer/topic_21_mirrorTree/Solution.java
题目描述 :请完成一个函数,输入一个二叉树,该函数输出它的镜像。
例如输入:
4
/ \
2 7
/ \ / \
1 3 6 9
镜像输出:
4
/ \
7 2
/ \ / \
9 6 3 1
示例 1:
输入:root = [4,2,7,1,3,6,9]
输出:[4,7,2,9,6,3,1]
限制:0 <= 节点个数 <= 1000
二叉树镜像定义: 对于二叉树中任意节点 root ,设其左 / 右子节点分别为 left, right;则在二叉树的镜像中的对应 root节点,其左 / 右子节点分别为 right, left 。
“Q:为何需要暂存root的左子节点? A:在递归右子节 点“root.left = mirrorTree(root.right);"执行完毕后,root.left 的值已经发生改变,此时递归左子节点mirrorTree(root.left)则会出问题。
复杂度分析:
package com.nateshao.sword_offer.topic_21_mirrorTree;
import java.util.Stack;
/**
* @date Created by 邵桐杰 on 2021/11/24 22:48
* @微信公众号 程序员千羽
* @个人网站 www.nateshao.cn
* @博客 https://nateshao.gitee.io
* @GitHub https://github.com/nateshao
* @Gitee https://gitee.com/nateshao
* Description: 剑指 Offer 27. 二叉树的镜像
*/
public class Solution {
/**
* 递归
*
* @param root
* @return
*/
public TreeNode mirrorTree(TreeNode root) {
if (root == null) return null;
TreeNode node = root.left;
root.left = mirrorTree(root.right);
root.right = mirrorTree(node);
return root;
}
/**
* 解法一:递归,时间复杂度:O(n),空间复杂度:O(n)
*
* @param root
* @return
*/
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return isMirror(root.left, root.right);
}
private boolean isMirror(TreeNode leftNode, TreeNode rightNode) {
if (leftNode == null && rightNode == null) return true;
if (leftNode == null || rightNode == null) return false;
return leftNode.val == rightNode.val && isMirror(leftNode.left, rightNode.right) && isMirror(leftNode.right, rightNode.left);
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
}
算法流程:
复杂度分析:
package com.nateshao.sword_offer.topic_21_mirrorTree;
import java.util.Stack;
/**
* @date Created by 邵桐杰 on 2021/11/24 22:48
* @微信公众号 程序员千羽
* @个人网站 www.nateshao.cn
* @博客 https://nateshao.gitee.io
* @GitHub https://github.com/nateshao
* @Gitee https://gitee.com/nateshao
* Description: 剑指 Offer 27. 二叉树的镜像
*/
public class Solution {
/**
* 栈
*
* @param root
* @return
*/
public TreeNode mirrorTree1(TreeNode root) {
if (root == null) return null;
Stack<TreeNode> stack = new Stack<TreeNode>() {{
add(root);
}};
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node.left != null) stack.add(node.left);
if (node.right != null) stack.add(node.right);
TreeNode tmp = node.left;
node.left = node.right;
node.right = tmp;
}
return root;
}
/**
* 解法二:迭代,时间复杂度:O(n),空间复杂度:O(n)
*
* @param root
* @return
*/
public boolean isSymmetric2(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
stack.push(root);
while (!stack.isEmpty()) {
TreeNode t1 = stack.pop();
TreeNode t2 = stack.pop();
if (t1 == null && t2 == null) continue;
if (t1 == null || t2 == null) return false;
if (t1.val != t2.val) return false;
stack.push(t1.left);
stack.push(t2.right);
stack.push(t1.right);
stack.push(t2.left);
}
return true;
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
}
参考链接:https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/solution/mian-shi-ti-27-er-cha-shu-de-jing-xiang-di-gui-fu-
革命尚未成功,同志仍需努力,冲冲冲