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。
栈
// 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;
}
}