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?
用栈或者Morris遍历。
{% if book.java %}
// 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;
}
}{% endif %}
{% if book.cpp %}
// 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;
}
};{% endif %}
{% codesnippet "./code/binary-tree-postorder-traversal-2."+book.suffix, language=book.suffix %}{% endcodesnippet %}