解法1

通过递归的方式:

class Solution {public boolean isSymmetric(TreeNode root) {if(null == root) {return false;}return check(root.left,root.right);}private boolean check(TreeNode left, TreeNode right) {if(null == left && null == right) {return true;}else if (null != left && null != right) {if(left.val == right.val) {return check(left.left, right.right) && check(left.right, right.left);}else {return false;}}return false;}
}

解法2

可以使用层次遍历,使用2个队列存在左子树,右子树. 同时在放入右子树的时候需要变换一下右子树的子节点的位置.

class Solution {public boolean isSymmetric(TreeNode root) {if(null == root) {return false;}LinkedList<TreeNode> leftQueue = new LinkedList<>();LinkedList<TreeNode> rightQueue = new LinkedList<>();leftQueue.offer(root.left);rightQueue.offer(root.right);while (! leftQueue.isEmpty() && ! rightQueue.isEmpty()) {TreeNode l = leftQueue.poll();TreeNode r = rightQueue.poll();if(l == null && r == null) {continue;}else if(l != null && r != null && l.val == r.val) {leftQueue.offer(l.left);leftQueue.offer(l.right);rightQueue.offer(r.right);rightQueue.offer(r.left);}else {return false;}}return leftQueue.isEmpty() && rightQueue.isEmpty();}}

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101.symmetric-tree

解法1

解法2

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