226. 翻转二叉树 101. 对称二叉树

226. 翻转二叉树

dfs:

/**
 * 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 {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) return null;
        f(root);
        return root;
    }
    void f(TreeNode root){
        if(root == null) return;
        if(root.left != null && root.right != null){
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        }else if(root.left == null && root.right != null){
            root.left = root.right;
            root.right = null;
        }else if(root.left != null && root.right == null){
            root.right = root.left;
            root.left = null;
        }else{
            return;
        }
        f(root.left);
        f(root.right);
    }
}

去掉代码冗余：

/**
 * 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 {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) return null;
        f(root);
        return root;
    }
    void f(TreeNode root){
        if(root == null) return;
       
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        
        f(root.left);
        f(root.right);
    }
}

101. 对称二叉树

此题目使用递归最简单

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null||(root.left==null&&root.right==null)) return true;
        return f(root.left,root.right);
    }
    public boolean f(TreeNode A,TreeNode B){
        if(A ==null && B ==null){
            return true;
        }else if(A !=null && B ==null){
            return false;
        }else if(A ==null && B !=null){
            return false;
        }else if(A.val != B.val){
            return false;
        }
        return f(A.left,B.right) && f(A.right,B.left);
    }
}

使用队列迭代法

/**
     * 迭代法
     * 使用普通队列
     */
    public boolean isSymmetric3(TreeNode root) {
        Queue<TreeNode> deque = new LinkedList<>();
        deque.offer(root.left);
        deque.offer(root.right);
        while (!deque.isEmpty()) {
            TreeNode leftNode = deque.poll();
            TreeNode rightNode = deque.poll();
            if (leftNode == null && rightNode == null) {
                continue;
            }
//            if (leftNode == null && rightNode != null) {
//                return false;
//            }
//            if (leftNode != null && rightNode == null) {
//                return false;
//            }
//            if (leftNode.val != rightNode.val) {
//                return false;
//            }
            // 以上三个判断条件合并
            if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
                return false;
            }
            // 这里顺序与使用Deque不同
            deque.offer(leftNode.left);
            deque.offer(rightNode.right);
            deque.offer(leftNode.right);
            deque.offer(rightNode.left);
        }
        return true;
    }