深度优先搜索
class Solution {
public boolean isSymmetric(TreeNode root) {
return isSymmetric(root.left, root.right);
}
/**
* 后序遍历
* 从根节点的左右孩子开始递归
*/
public boolean isSymmetric(TreeNode left, TreeNode right){
/**
* 除非全是null或全不是null,否则返回false
*/
if (left == null || right == null){
return left == right;
}
if (left.val != right.val){
return false;
}
return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
}
/**
* 时间复杂度 O(n)
* 空间复杂度 O(logn)
*/
迭代(队列)
class Solution {
public boolean isSymmetric(TreeNode root) {
/**
* 迭代,使用队列的先进先出原则
* 先将根节点的左右孩子入队,如果满足条件,再将左孩子的左孩子和右孩子的右孩子、左孩子的右孩子和右孩子的左孩子入队
* 每次出队两个节点进行比较
*/
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root.left);
queue.add(root.right);
while (!queue.isEmpty()){
TreeNode left = queue.poll();
TreeNode right = queue.poll();
if (left == null && right == null){
continue;
}
if (left == null || right == null){
return false;
}
if (left.val != right.val){
return false;
}
queue.add(left.left);
queue.add(right.right);
queue.add(left.right);
queue.add(right.left);
}
return true;
}
}
/**
* 时间复杂度 O(n)
* 空间复杂度 O(n)
*/
迭代(栈)
class Solution {
public boolean isSymmetric(TreeNode root) {
/**
* 在本题中,使用栈和队列是一模一样的
*/
Stack<TreeNode> stack = new Stack<>();
stack.push(root.left);
stack.push(root.right);
while (!stack.isEmpty()){
TreeNode left = stack.pop();
TreeNode right = stack.pop();
if (left == null && right == null){
continue;
}
if (left == null || right == null){
return false;
}
if (left.val != right.val){
return false;
}
stack.push(left.left);
stack.push(right.right);
stack.push(left.right);
stack.push(right.left);
}
return true;
}
}
https://leetcode-cn.com/problems/symmetric-tree/