学完层序,真的直接一口气,不用任何递归,搞定10道题

102. 二叉树的层序遍历

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> listList = new ArrayList<>();
        TreeNode cur = root;
        if (cur == null) {
            return listList;
        }
        queue.offer(cur);
        int len;
        while (!queue.isEmpty()) {
            len = queue.size();
            List<Integer> list = new ArrayList<>();
            while (len > 0) {
                cur = queue.poll();
                list.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                len--;
            }
            listList.add(list);
        }
        return listList;
    }
}

226. 翻转二叉树

class Solution {
    public TreeNode invertTree(TreeNode root) {
    Queue<TreeNode> queue = new LinkedList<>();
        TreeNode cur = root;
        if (cur == null) {
            return cur;
        }
        queue.offer(cur);
        int len;
        while (!queue.isEmpty()) {
            len = queue.size();
            int size = len;
            while (len > 0) {
                cur = queue.poll();
                swap(cur);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                len--;
            }
        }
        return root;
    }
    public void swap(TreeNode treeNode) {
        TreeNode node;
        node = treeNode.left;
        treeNode.left = treeNode.right;
        treeNode.right = node;
    }
}

101. 对称二叉树

class Solution {
   public boolean isSymmetric(TreeNode root) {
        return compare(root.left,root.right );
    }

    public boolean compare(TreeNode left, TreeNode right) {
        boolean bool1=false;
        if (left == null && right == null) {
            return true;
        }
        // 有了上面的判断, 就可以保证可不能是两个null,
        // 这样下面,只要有null,就一定是只有一个null
        if (left == null || right == null) {
            return false;
        }
        if (left.val == right.val) {
            bool1 = true;
        }
        boolean bool2 = compare(left.left, right.right);
        boolean bool3 = compare(left.right, right.left);
        return bool1 & bool2 & bool3;
    }
}