代码随想录算法训练营第十五天| 层序遍历、226.翻转二叉树、101. 对称二叉树

层序遍历

题目链接：. - 力扣（LeetCode）

解题思路：深度优先遍历用deep标记层数，广度优先遍历用队列实现

java：

class Solution {
    public List> resList = new ArrayList>();
    public List> levelOrder(TreeNode root) {
        //checkFun01(root,0);
        checkFun02(root);
        return resList;
    }
    //DFS--递归方式
    public void checkFun01(TreeNode node, Integer deep) {
        if (node == null) return;
        deep++;
        if (resList.size() < deep) {
            List item = new ArrayList();
            resList.add(item);
        }
        resList.get(deep - 1).add(node.val);
        checkFun01(node.left, deep);
        checkFun01(node.right, deep);
    }
    //BFS--迭代方式--借助队列
    public void checkFun02(TreeNode node) {
        if (node == null) return;
        Queue que = new LinkedList();
        que.offer(node);
        while (!que.isEmpty()) {
            List itemList = new ArrayList();
            int len = que.size();
            while (len > 0) {
                TreeNode tmpNode = que.poll();
                itemList.add(tmpNode.val);
                if (tmpNode.left != null) que.offer(tmpNode.left);
                if (tmpNode.right != null) que.offer(tmpNode.right);
                len--;
            }
            resList.add(itemList);
        }
    }
}

226.翻转二叉树

题目链接：. - 力扣（LeetCode）

解题思路：直接用递归交换左右结点，或者借助栈，注意不能中序遍历

C：

//递归
struct TreeNode* invertTree(struct TreeNode* root) {
    if(!root)
        return NULL;
    struct TreeNode* temp = root->right;
    root->right = root->left;
    root->left = temp;
    invertTree(root->left);
    invertTree(root->right);
    return root;
}
//栈
struct TreeNode* invertTree(struct TreeNode* root) {
    if(!root)
        return NULL;
    //存储结点的栈
    struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 100);
    int stackTop = 0;
    stack[stackTop++] = root;
    while(stackTop) {
        struct TreeNode* temp = stack[--stackTop];
        struct TreeNode* tempNode = temp->right;
        temp->right = temp->left;
        temp->left = tempNode;
        if(temp->right)
            stack[stackTop++] = temp->right;
        if(temp->left)
            stack[stackTop++] = temp->left;
    }
    return root;
}

java：

//DFS递归
class Solution {
   /**
     * 前后序遍历都可以
     * 中序不行，因为先左孩子交换孩子，再根交换孩子（做完后，右孩子已经变成了原来的左孩子），再右孩子交换孩子（此时其实是对原来的左孩子做交换）
     */
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        invertTree(root.left);
        invertTree(root.right);
        swapChildren(root);
        return root;
    }

    private void swapChildren(TreeNode root) {
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
    }
}

//BFS
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {return null;}
        ArrayDeque deque = new ArrayDeque<>();
        deque.offer(root);
        while (!deque.isEmpty()) {
            int size = deque.size();
            while (size-- > 0) {
                TreeNode node = deque.poll();
                swap(node);
                if (node.left != null) deque.offer(node.left);
                if (node.right != null) deque.offer(node.right);
            }
        }
        return root;
    }

    public void swap(TreeNode root) {
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
    }
}

101. 对称二叉树

题目链接：. - 力扣（LeetCode）

解题思路：只需要将left.left与right.right比较外侧，left.right与right.left比较内侧即可

class Solution {
    public boolean isSymmetric1(TreeNode root) {
        return compare(root.left, root.right);
    }
    private boolean compare(TreeNode left, TreeNode right) {
        if (left == null && right != null) {
            return false;
        }
        if (left != null && right == null) {
            return false;
        }
        if (left == null && right == null) {
            return true;
        }
        if (left.val != right.val) {
            return false;
        }
        boolean compareOutside = compare(left.left, right.right);
        boolean compareInside = compare(left.right, right.left);
        return compareOutside && compareInside;
    }
}