---------------🎈🎈对称二叉树 题目链接🎈🎈-------------------
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解法一: 递归:后序遍历 左右中
时间复杂度O(N)
空间复杂度O(N)文章来源地址https://www.toymoban.com/news/detail-825885.html
/**
* 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;
* }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
// 递归
return compare(root.left, root.right);
}
public boolean compare(TreeNode left, TreeNode right){ // 确定递归的参数和返回值
if(left == null && right==null){return true;}
if(left != null && right==null){return false;}
if(left == null && right!=null){return false;}
if(left.val != right.val){return false;}
// 递归逻辑:继续比较左右两个子树的内外侧【相当于后序遍历,最后返回内侧和外侧的比较结果】
boolean compareOutside = compare(left.left, right.right);
boolean compareInside = compare(left.right, right.left);
return compareInside && compareOutside; // 内外侧都是true的时候就返回true
}
}
解法二: 迭代法,用了单端队列
/**
* 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;
* }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
// 采用迭代法:用了单端队列
Queue<TreeNode> myqueue = new LinkedList<>();
myqueue.add(root.left);
myqueue.add(root.right);
while(!myqueue.isEmpty()){
TreeNode leftnode = myqueue.poll();
TreeNode rightnode = myqueue.poll();
if(leftnode == null && rightnode == null){
continue;
}
if(leftnode != null && rightnode == null){
return false;
}
if(leftnode == null && rightnode != null){
return false;
}
if(leftnode.val != rightnode.val){
return false;
}
myqueue.add(leftnode.left);
myqueue.add(rightnode.right);
myqueue.add(leftnode.right);
myqueue.add(rightnode.left);
}
return true;
}
}
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