Binary Tree Postorder Traversal
描述
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
分析
用栈或者Morris遍历。
栈
// Binary Tree Postorder Traversal
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> result = new ArrayList<>();
Stack<TreeNode> s = new Stack<>();
/* p，正在访问的结点，q，刚刚访问过的结点*/
TreeNode p = root;
TreeNode q = null;
do {
while (p != null) { /* 往左下走*/
s.push(p);
p = p.left;
}
q = null;
while (!s.empty()) {
p = s.pop();
/* 右孩子不存在或已被访问，访问之*/
if (p.right == q) {
result.add(p.val);
q = p; /* 保存刚访问过的结点*/
} else {
/* 当前结点不能访问，需第二次进栈*/
s.push(p);
/* 先处理右子树*/
p = p.right;
break;
}
}
} while (!s.empty());
return result;
}
}
Morris后序遍历
// Binary Tree Postorder Traversal
// Morris后序遍历，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public List postorderTraversal(TreeNode root) {
ArrayList result = new ArrayList<>();
TreeNode dummy = new TreeNode(-1);
dummy.left = root;
TreeNode cur = dummy;
TreeNode prev = null;
while (cur != null) {
if (cur.left == null) {
prev = cur; /* 必须要有 */
cur = cur.right;
} else {
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; /* 必须要有 */
cur = cur.left;
} else { /* 已经线索化，则访问节点，并删除线索 */
visit_reverse(cur.left, prev, result);
prev.right = null;
prev = cur; /* 必须要有 */
cur = cur.right;
}
}
}
return result;
}
// 逆转路径
private static void reverse(TreeNode from, TreeNode to) {
TreeNode x = from;
TreeNode y = from.right;
TreeNode z = null;
if (from == to) return;
while (x != to) {
z = y.right;
y.right = x;
x = y;
y = z;
}
}
// 访问逆转后的路径上的所有结点
private static void visit_reverse(TreeNode from, TreeNode to,
List result) {
TreeNode p = to;
reverse(from, to);
while (true) {
result.add(p.val);
if (p == from)
break;
p = p.right;
}
reverse(to, from);
}
}