LeetCode 101-105

101. 对称二叉树

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return check(root.left, root.right);
    }

    public boolean check(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null || left.val != right.val) {
            return false;
        }
        return check(left.left, right.right) && check(left.right, right.left);
    }
}

102. 二叉树的层序遍历

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

103. 二叉树的锯齿形层次遍历

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }
        List<List<Integer>> res = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        boolean flag = true;//true正向，false反向
        while (!queue.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode front = queue.poll();
                list.add(front.val);
                if (front.left != null) {
                    queue.offer(front.left);
                }
                if (front.right != null) {
                    queue.offer(front.right);
                }
            }
            if (flag == true) {
                res.add(new ArrayList<>(list));
                flag = false;
            } else {
                Collections.reverse(list);
                res.add(new ArrayList<>(list));
                flag = true;
            }
        }
        return res;
    }
}

104. 二叉树的最大深度

class Solution {
    int res = 0;

    public void dfs(int level, TreeNode root) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right == null) {
            res = Math.max(res, level);
        }
        dfs(level + 1, root.left);
        dfs(level + 1, root.right);
    }

    public int maxDepth(TreeNode root) {
        dfs(1, root);
        return res;
    }
}

105. 从前序与中序遍历序列构造二叉树

class Solution {
    public TreeNode build(int preL, int preR, int inL, int inR, int[] preOrder, int[] inOrder) {
        if (preL > preR) {
            return null;
        }
        int rootVal = preOrder[preL];
        TreeNode root = new TreeNode(rootVal);
        int index;
        for (index = inL; index <= inR; index++) {
            if (inOrder[index] == rootVal) {
                break;
            }
        }
        int leftNum = index - inL;
        root.left = build(preL + 1, preL + leftNum, inL, inL + leftNum, preOrder, inOrder);
        root.right = build(preL + leftNum + 1, preR, index + 1, inR, preOrder, inOrder);
        return root;

    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return build(0, preorder.length - 1, 0, inorder.length - 1, preorder, inorder);
    }
}