Binary Tree Inorder Traversal
描述
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
分析
用栈或者Morris遍历。
栈
// Binary Tree Inorder Traversal
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
ArrayList<Integer> result = new ArrayList<>();
Stack<TreeNode> s = new Stack<>();
TreeNode p = root;
while (!s.empty() || p != null) {
if (p != null) {
s.push(p);
p = p.left;
} else {
p = s.pop();
result.add(p.val);
p = p.right;
}
}
return result;
}
}
Morris中序遍历
// Binary Tree Inorder Traversal
// Morris中序遍历，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public List inorderTraversal(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 = 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 */
node.right = cur;
/* prev = cur; 不能有这句，cur还没有被访问 */
cur = cur.left;
} else { /* 已经线索化，则访问节点，并删除线索 */
result.add(cur.val);
node.right = null;
prev = cur;
cur = cur.right;
}
}
}
return result;
}
}