[leetCode]剑指 Offer 27. 二叉树的镜像

递归

通过观察可以发现，只需交换非叶子节点的左右子节点即可完成镜像。
通过前序遍历，从上至下进行递归，如果非叶子节点则交换其子节点，如果为叶子节点，则返回其本身（只有一个节点的情况）。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root ==  null) return null;
        if(root.left == null && root.right == null) return root;
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        if(root.left != null)
            mirrorTree(root.left);
        if(root.right != null)
            mirrorTree(root.right);
        return root;
    }
}

辅助栈

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root == null) return null;
        LinkedList<TreeNode> stack = new LinkedList<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            if(node.left != null ) stack.push(node.left);
            if(node.right != null) stack.push(node.right); 
            if(node.left == null && node.right == null) continue;//跳过叶子节点
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;
        }
        return root;
    }
}