101. 对称二叉树 + 递归 + 迭代
101. 对称二叉树
LeetCode_101
题目描述
方法一:递归法
/**
* 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) {
if(root == null)
return false;
return isMirror(root.left, root.right);
}
public boolean isMirror(TreeNode left, TreeNode right){
if(left == null && right == null)
return true;
if(left == null || right == null)
return false;
if(left.val == right.val)//左子树的左节点和右子树的右节点是镜像对称的而且左子树的右节点和左子树的左节点是镜像对称的。
return isMirror(left.left, right.right) && isMirror(left.right, right.left);
else return false;
}
}
方法二:迭代法
/**
* 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) {
if(root == null)
return false;
Queue<TreeNode> que = new LinkedList<>();
que.offer(root.left);
que.offer(root.right);
while(!que.isEmpty()){
TreeNode left = que.poll();
TreeNode right = que.poll();
if(left == null && right == null)
continue;
if(left == null || right == null || left.val != right.val)
return false;
que.offer(left.left);
que.offer(right.right);
que.offer(left.right);
que.offer(right.left);
}
return true;
}
}
Either Excellent or Rusty