Description
Given the root of a binary tree, return the preorder traversal of its nodes’ values.
Example 1:
Input: root = [1,null,2,3] Output: [1,2,3]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
Code
Iterative
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
TreeNode* p = root;
while(p || !st.empty()) {
if(p) {
res.push_back(p->val);
st.push(p);
p = p->left;
} else {
p = st.top(); st.pop();
p = p->right;
}
}
return res;
}
};將所有 node 都 push 到 stack 上，經由 pop 的過程來執行 preorder traversal。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> nodes;
stack<TreeNode*> st;
if(!root) return {};
st.push(root);
while(!st.empty()) {
auto node = st.top(); st.pop();
nodes.push_back(node->val);
if(node->right) st.push(node->right);
if(node->left) st.push(node->left);
}
return nodes;
}
};另一種做法是只將 right child push 到 stack 上，其他部分的 traversal 藉由 root pointer 的移動來達成。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> nodes;
stack<TreeNode*> st;
while(root || !st.empty()) {
if(root) {
nodes.push_back(root->val);
if(root->right) {
st.push(root->right);
}
root = root->left;
} else {
root = st.top();
st.pop();
}
}
return nodes;
}
};