力扣翻转二叉树的四种解法

翻转二叉树

解法如下：

/**
 * 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 {
    //前序遍历方法
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
    //中序遍历方法
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        invertTree(root.left);
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        // 因为中序遍历已经将左右子树的位置互换，所以要传入实际的右子节点，即root.left
        invertTree(root.left);
        return root;
    }
    //后序遍历方法
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        invertTree(root.left);
        invertTree(root.right);
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        return root;
    }
    //层序遍历方法
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;
            if (node.left != null) queue.offer(node.left);
            if (node.right != null) queue.offer(node.right);
        }
        return root;
    }
}