面试题32 - III. 从上到下打印二叉树 III

请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。

例如:
给定二叉树: [3,9,20,null,null,15,7],

返回其层次遍历结果：

[
  [3],
  [20,9],
  [15,7]
]

提示：

节点总数 <= 1000

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

两个栈

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) {
            return res;
        }

        Stack<TreeNode> oddstack = new Stack<>();
        Stack<TreeNode> evenstack = new Stack<>();

        List<Integer> list;

        oddstack.add(root);
        while (!oddstack.isEmpty() || !evenstack.isEmpty()) {
            list = new ArrayList<>();
            while (!oddstack.isEmpty()) {
                TreeNode node = oddstack.pop();
                list.add(node.val);
                TreeNode leftnode = node.left;
                if(leftnode != null) {
                    evenstack.add(leftnode);
                }
                TreeNode rightnode = node.right;
                if(rightnode != null) {
                    evenstack.add(rightnode);
                }
            }
            res.add(list);

            list = new ArrayList<>();
            while (!evenstack.isEmpty()) {
                TreeNode node = evenstack.pop();
                list.add(node.val);
                TreeNode rightnode = node.right;
                if(rightnode != null) {
                    oddstack.add(rightnode);
                }
                TreeNode leftnode = node.left;
                if(leftnode != null) {
                    oddstack.add(leftnode);
                }
            }
            if(list.size() != 0) {
                res.add(list);
            }
        }
        return res;
    }

参考

    List<List<Integer>> res = new ArrayList<>();

    public List<List<Integer>> levelOrder(TreeNode root) {
        helper(root, 0);
        return res;
    }

    private void helper(TreeNode root, int level) {
        if (root == null)
            return;
        if (res.size() == level) {
            res.add(new ArrayList<>());
        }
        if (level % 2 == 0) {
            res.get(level).add(root.val);
        } else {
            // 加在最前面
            res.get(level).add(0, root.val);
        }
        helper(root.left, level + 1);
        helper(root.right, level + 1);
    }

官方题解

class Solution {
    public String convert(String s, int numRows) {

        if (numRows == 1) return s;

        List<StringBuilder> rows = new ArrayList<>();
        for (int i = 0; i < Math.min(numRows, s.length()); i++)
            rows.add(new StringBuilder());

        int curRow = 0;
        boolean goingDown = false;

        for (char c : s.toCharArray()) {
            rows.get(curRow).append(c);
            if (curRow == 0 || curRow == numRows - 1) goingDown = !goingDown;
            curRow += goingDown ? 1 : -1;
        }

        StringBuilder ret = new StringBuilder();
        for (StringBuilder row : rows) ret.append(row);
        return ret.toString();
    }
}

作者：LeetCode
链接：https://leetcode-cn.com/problems/zigzag-conversion/solution/z-zi-xing-bian-huan-by-leetcode/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

巴特西

面试题32 - III. 从上到下打印二叉树 III

面试题32 - III. 从上到下打印二叉树 III

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