二叉树是常见的数据结构,二叉树相关的算法题目也是非常常见的。下面总结以下二叉树的前序/中序/后序遍历方法,分别用递归(O(n) Space),迭代(O(n) Space),Morris(O(1) Space)方法实现。
二叉树遍历的时间复杂度都是O(n),不同方法的区别主要是在空间复杂度上。
Recursive Traversal
递归是一种非常直观的方法,也是最容易实现的方法。递归的方法的空间复杂度为O(n)。
节点的定义如下:
// Definition for a binary tree node.
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}
Preorder Traversal
void preorderTraversal(TreeNode* root, vector<int>& nums) {
if(!root) return;
nums.push_back(root->val);
preorderTraversal(root->left);
preorderTraversal(root->right);
}
Inorder Traversal
void inorderTraversal(TreeNode* root, vector<int>& nums) {
if(!root) return;
inorderTraversal(root->left, nums);
nums.push_back(root->val);
inorderTraversal(root->right, nums);
}
Postorder Traversal
void postorderTraversal(TreeNode* root, vector<int>& nums) {
if(!root) return;
postorderTraversal(root->left, nums);
postorderTraversal(root->right, nums);
nums.push_back(root->val);
}
Iterative Traversal
迭代方法需要使用stack来保存遍历路径上待遍历的节点,从根节点到叶节点,最多保存n/2个节点。空间复杂度为O(n)。
Preorder Traversal
void preorderTraversal(TreeNode* root, vector<int>& nums) {
vector<int> nums;
vector<int> nums;
stack<TreeNode* > st; // pointers in stack are valid
while (root || !st.empty()) {
if (!root) {
root = st.top();
st.pop();
}
nums.push_back(root->val);
// if valid right child, push into stack
if (root->right) st.push(root->right);
// go to the left child
root = root->left;
}
return nums;
}
Inorder Traversal
vector<int> inorderTraversal(TreeNode* root) {
vector<int> nums;
stack<TreeNode* > st; // pointers in stack are valid
while (root || !st.empty()) {
if (root) {
// push root into stack, then go left
st.push(root);
root = root->left;
} else {
// for nodes in stack, only visit its right
root = st.top();
st.pop();
nums.push_back(root->val);
root = root->right;
}
}
return nums;
}
Postorder Traversal
后序遍历可以看作是和前序遍历是左右对称的,从根节点开始,先遍历右子树,再遍历左子树,只不过为了得到后序遍历的输出,我们需要遍历结果逆序输出。可以比较前序遍历的代码,逻辑完全是一样的,就是左右子树访问顺序交换了,完全是对称的。
vector<int> postorderTraversal(TreeNode* root) {
vector<int> nums;
stack<TreeNode* > stnode;
while (root || !stnode.empty()) {
if (!root) {
root = stnode.top();
stnode.pop();
}
nums.push_back(root->val);
if (root->left) stnode.push(root->left);
root = root->right;
}
return vector<int>(nums.rbegin(), nums.rend())
}
Morris Traversal
Morris遍历方法不需要栈来保存待访问的节点,而是通过利用节点本身的指针来保存待访问节点的指针,并在访问过程中恢复节点。实现了O(1)空间复杂度。
Preorder Traversal
图片摘自
Reference4
vector<int> preorderTraversal(TreeNode* root) {
// morris traversal
vector<int> nums;
TreeNode* cur = nullptr;
while (root) {
if (root->left) {
cur = root->left;
// find the predecessor of root node
while (cur->right && cur->right != root) {
cur = cur->right;
}
// has visited this root node
if (cur->right == root) {
cur->right = nullptr;
root = root->right;
} else {
nums.push_back(root->val);
cur->right = root;
root = root->left;
}
} else {
nums.push_back(root->val);
root = root->right;
}
}
return nums;
}
Inorder Traversal
中序和前序代码基本一样,唯一不同的在输出节点值的顺序不同。
图片摘自
Reference4
vector<int> inorderTraversal(TreeNode* root) {
vector<int> inorderTraversal(TreeNode* root) {
vector<int> nums;
TreeNode* cur = nullptr;
while (root) {
if (root->left) {
cur = root->left;
while (cur->right && cur->right != root) {
cur = cur->right;
}
if (cur->right == root) {
nums.push_back(root->val);
cur->right = nullptr;
root = root->right;
} else {
cur->right = root;
root = root->left;
}
} else {
nums.push_back(root->val);
root = root->right;
}
}
return nums;
}
}
Postorder Traversal
之前我们在讲迭代方法时,说过了后序遍历其实可以看作是和前序遍历左右对称的,此处,我们同样可以利用这个性质,基于前序遍历的算法,可以很快得到后序遍历的结果。我们只需要将前序遍历中所有的左孩子和右孩子进行交换就可以了。
图片摘自
Reference4
vector<int> postorderTraversal(TreeNode* root) {
vector<int> nums;
TreeNode* cur = nullptr;
while (root) {
if (root->right) {
cur = root->right;
while (cur->left && cur->left != root) {
cur = cur->left;
}
if (cur->left == root) {
cur->left = nullptr;
root = root->left;
} else {
nums.push_back(root->val);
cur->left = root;
root = root->right;
}
} else {
nums.push_back(root->val);
root = root->left;
}
}
return vector<int>(nums.rbegin(), nums.rend());
}
Reference
1. https://leetcode.com/problems/binary-tree-preorder-traversal/description/
2. https://leetcode.com/problems/binary-tree-inorder-traversal/description/
3. https://leetcode.com/problems/binary-tree-postorder-traversal/description/
4. http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html