144.Binary Tree Preorder Traversal
Given the root of a binary tree, return the preorder traversal of its nodes’ values.
Example 1:
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| Input: root = [1,null,2,3]
Output: [1,2,3]
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Example 2:
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| Input: root = []
Output: []
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Example 3:
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| Input: root = [1]
Output: [1]
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时间复杂度：O(n)
空间复杂度：O(n)
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| class Solution {
public:
void preorder(TreeNode *root, vector<int> &ans_v) {
if (root == nullptr)return;
else {
ans_v.push_back(root->val);
preorder(root->left,ans_v);
preorder(root->right,ans_v);
}
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> ans_v;
preorder(root,ans_v);
return ans_v;
}
};
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迭代法：
时间复杂度：O(n)
空间复杂度：O(n)
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| class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
if (root == nullptr) {
return res;
}
stack<TreeNode*> stk;
TreeNode* node = root;
while (!stk.empty() || node != nullptr) {
while (node != nullptr) {
res.emplace_back(node->val);
stk.emplace(node);
node = node->left;
}
node = stk.top();
stk.pop();
node = node->right;
}
return res;
}
};
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Morris 遍历：
时间复杂度：O(n)
空间复杂度：O(1)
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| class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> res;
if (root == nullptr) {
return res;
}
TreeNode *p1 = root, *p2 = nullptr;
while (p1 != nullptr) {
p2 = p1->left;
if (p2 != nullptr) {
while (p2->right != nullptr && p2->right != p1) {
p2 = p2->right;
}
if (p2->right == nullptr) {
res.emplace_back(p1->val);
p2->right = p1;
p1 = p1->left;
continue;
} else {
p2->right = nullptr;
}
} else {
res.emplace_back(p1->val);
}
p1 = p1->right;
}
return res;
}
};
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