binary-tree/traversal/binary-tree-postorder-traversal
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2023-12-01
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;
}
}
// Binary Tree Postorder Traversal
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
stack<const TreeNode *> s;
/* p，正在访问的结点，q，刚刚访问过的结点*/
const TreeNode *p = root, *q = nullptr;
do {
while (p != nullptr) { /* 往左下走*/
s.push(p);
p = p->left;
}
q = nullptr;
while (!s.empty()) {
p = s.top();
s.pop();
/* 右孩子不存在或已被访问，访问之*/
if (p->right == q) {
result.push_back(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<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> 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<Integer> result) {
TreeNode p = to;
reverse(from, to);
while (true) {
result.add(p.val);
if (p == from)
break;
p = p.right;
}
reverse(to, from);
}
}
// 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:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
TreeNode dummy(-1);
TreeNode *cur, *prev = nullptr;
std::function < void(const TreeNode*)> visit =
[&result](const TreeNode *node){
result.push_back(node->val);
};
dummy.left = root;
cur = &dummy;
while (cur != nullptr) {
if (cur->left == nullptr) {
prev = cur; /* 必须要有 */
cur = cur->right;
} else {
TreeNode *node = cur->left;
while (node->right != nullptr && node->right != cur)
node = node->right;
if (node->right == nullptr) { /* 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, visit);
prev->right = nullptr;
prev = cur; /* 必须要有 */
cur = cur->right;
}
}
}
return result;
}
private:
// 逆转路径
static void reverse(TreeNode *from, TreeNode *to) {
TreeNode *x = from, *y = from->right, *z;
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
static void visit_reverse(TreeNode* from, TreeNode *to,
std::function< void(const TreeNode*) >& visit) {
TreeNode *p = to;
reverse(from, to);
while (true) {
visit(p);
if (p == from)
break;
p = p->right;
}
reverse(to, from);
}
};