Binary Tree Preorder Traversal

描述

Given a binary tree, return the preorder traversal of its nodes' values.

For example: Given binary tree {1,#,2,3},

 1
  \
   2
  /
 3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

分析

用栈或者Morris遍历。

// Binary Tree Preorder Traversal
// 使用栈,时间复杂度O(n),空间复杂度O(n)
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<>();
        Stack<TreeNode> s = new Stack<>();
        if (root != null) s.push(root);

        while (!s.isEmpty()) {
            final TreeNode p = s.pop();
            result.add(p.val);

            if (p.right != null) s.push(p.right);
            if (p.left != null) s.push(p.left);
        }
        return result;
    }
}

Morris先序遍历

// Binary Tree Preorder Traversal
// Morris先序遍历,时间复杂度O(n),空间复杂度O(1)
class Solution {
    public List preorderTraversal(TreeNode root) {
        ArrayList result = new ArrayList<>();
        TreeNode cur = root;
        TreeNode prev = null;

        while (cur != null) {
            if (cur.left == null) {
                result.add(cur.val);
                prev = cur; /* cur刚刚被访问过 */
                cur = cur.right;
            } else {
                /* 查找前驱 */
                TreeNode node = cur.left;
                while (node.right != null && node.right != cur)
                    node = node.right;

                if (node.right == null) { /* 还没线索化,则建立线索 */
                    result.add(cur.val); /* 仅这一行的位置与中序不同 */
                    node.right = cur;
                    prev = cur; /* cur刚刚被访问过 */
                    cur = cur.left;
                } else {    /* 已经线索化,则删除线索  */
                    node.right = null;
                    /* prev = cur; 不能有这句,cur已经被访问 */
                    cur = cur.right;
                }
            }
        }
        return result;
    }
}

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