剑指Offer32-34题解

/**
 * https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/
 *
 * @author Qitong!!
 * @Date 2020/7/4
 */
public class 剑指Offer_32_I_从上到下打印二叉树 {
    //二叉树的层序遍历
    public int[] levelOrder(TreeNode root) {
        if (root == null) return new int[0];

        Queue<TreeNode> queue = new LinkedList<>();
        List<Integer> temp = new ArrayList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode curNode = queue.poll();
            temp.add(curNode.val);
            if (curNode.left != null) queue.offer(curNode.left);
            if (curNode.right != null) {
                queue.offer(curNode.right);
            }
        }
        int[] res = new int[temp.size()];
        int curI = 0;
        for (Integer integer : temp) {
            res[curI++] = integer;
        }
        return res;
    }
}

/**
 * https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/
 *
 * @author Qitong!!
 * @Date 2020/7/4
 */
public class 剑指Offer_32_II_从上到下打印二叉树II {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode curNode = queue.poll();
                temp.add(curNode.val);
                if (curNode.left != null) queue.offer(curNode.left);
                if (curNode.right != null) queue.offer(curNode.right);
            }
            res.add(temp);
        }
        return res;
    }
}

/**
 * https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/
 *
 * @author Qitong!!
 * @Date 2020/7/4
 */
public class 剑指Offer_32_III从上到下打印二叉树III {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        //增加标志位! flag为 true表示从右往左打印
        boolean flag = false;
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode curNode = queue.poll();
                temp.add(curNode.val);
                if (curNode.left != null) queue.offer(curNode.left);
                if (curNode.right != null) queue.offer(curNode.right);
            }
            if (flag) Collections.reverse(temp);
            flag = !flag;
            res.add(temp);
        }
        return res;
    }
}

/**
 * https://leetcode-cn.com/problems/er-cha-sou-suo-shu-de-hou-xu-bian-li-xu-lie-lcof/
 *
 * @author Qitong!!
 * @Date 2020/7/4
 */
public class 剑指Offer_33_二叉搜索树的后序遍历序列 {
    //递归做法!
    public boolean verifyPostorder(int[] postorder) {
        return process(postorder, 0, postorder.length - 1);
    }

    //[l,r]
    private boolean process(int[] postorder, int l, int r) {
        if (l >= r) return true;

        int p = l;
        while (postorder[p] < postorder[r]) p++;
        int m = p;
        while (postorder[p] > postorder[r]) p++;
        //根结点是否为最右边的!
        return p == r && process(postorder, l, m - 1) && process(postorder, m, r - 1);
    }
}

/**
 * https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/
 *
 * @author Qitong!!
 * @Date 2020/7/5
 */
public class 剑指Offer_34_二叉树中和为某一值的路径 {
    private List<List<Integer>> list;
    private List<Integer> res;

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        list = new ArrayList<>();
        if (root == null) return list;
        res = new ArrayList<>();

        dfs(root, sum);
        return list;
    }

    private void dfs(TreeNode root, int sum) {
        if (root == null) {
            if (sum == 0) {
                list.add(new ArrayList<>(res));
            }
            return;
        }

        res.add(root.val);
        if (root.left == null) {
            dfs(root.right, sum - root.val);
        } else if (root.right == null) {
            dfs(root.left, sum - root.val);
        } else {
            dfs(root.left, sum - root.val);
            dfs(root.right, sum - root.val);
        }
        res.remove(res.size() - 1);
    }
}