把三个二叉树遍历的题放在一起了。
递归写法太简单，就不再实现了，每题实现了两种非递归算法。
一种是利用栈，时间和空间复杂度都是O(n)。
另一种是借助线索二叉树，也叫Morris遍历，充分利用树中节点的空指针域。
先序：
Binary Tree Preorder Traversal
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].
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode *> s;
if (root) {
s.push(root);
}
while (!s.empty()) {
TreeNode *p = s.top();
s.pop();
result.push_back(p->val);
if (p->right) {
s.push(p->right);
}
if (p->left) {
s.push(p->left);
}
}
return result;
}
}; class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
TreeNode *p = root;
while (p) {
if (!p->left) {
result.push_back(p->val);
p = p->right;
} else {
TreeNode *q = p->left;
while (q->right && q->right != p) {
q = q->right;
}
if (q->right == nullptr) {
result.push_back(p->val);
q->right = p;
p = p->left;
} else {
q->right = nullptr;
p = p->right;
}
}
}
return result;
}
};中序：
Binary Tree Inorder Traversal
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].
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
vector<int> result;
TreeNode *p = root;
stack<TreeNode *> s;
while (!s.empty() || p != nullptr) {
if (p != nullptr) {
s.push(p);
p = p->left;
} else {
p = s.top();
s.pop();
result.push_back(p->val);
p = p->right;
}
}
return result;
}
}; class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
vector<int> result;
TreeNode *p = root;
while (p) {
if (p->left) {
TreeNode *q = p->left;
while (q->right && q->right != p) {
q = q->right;
}
if (q->right == p) {
result.push_back(p->val);
q->right = nullptr;
p = p->right;
} else {
q->right = p;
p = p->left;
}
} else {
result.push_back(p->val);
p = p->right;
}
}
return result;
}
};后序：
Binary Tree Postorder Traversal
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].
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode *> s;
TreeNode *p = root;
TreeNode *q = nullptr, *last = nullptr;
while (!s.empty() || p) {
if (p) {
s.push(p);
p = p->left;
} else {
q = s.top();
if (q->right && last != q->right) {
p = q->right;
} else {
result.push_back(q->val);
s.pop();
last = q;
}
}
}
return result;
}
}; class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
TreeNode dummy();
TreeNode *p = &dummy, *pre = nullptr;
p->left = root;
while (p) {
if (p->left == nullptr) {
p = p->right;
} else {
pre = p->left;
while (pre->right != nullptr && pre->right != p) {
pre = pre->right;
}
if (pre->right == nullptr) {
pre->right = p;
p = p->left;
} else {
printReverse(result, p->left, pre);
pre->right = nullptr;
p = p->right;
}
}
}
return result;
}
private:
void printReverse(vector<int>& v, TreeNode* from, TreeNode *to) {
reverse(from, to);
TreeNode *p = to;
while (true) {
v.push_back(p->val);
if (p == from) {
break;
}
p = p->right;
}
reverse(to, from);
}
void reverse(TreeNode *from, TreeNode *to) {
if (from == to) {
return;
}
TreeNode *x = from, *y = x->right, *z;
while (true) {
z = y->right;
y->right = x;
x = y;
y = z;
if (x == to) {
break;
}
}
}
};