94. Binary Tree Inorder Traversal(Tree, stack)
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
法I: recursion
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
inorder(root);
return ret;
} void inorder(TreeNode* root){
if(root==NULL) return; inorder(root->left);
ret.push_back(root->val);
inorder(root->right);
return; }
private:
vector<int> ret;
};
法II:iteration
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
if(root==NULL) return ret; TreeNode* current = root;
stack<TreeNode*> s;
s.push(current);
while(){
while(current->left){
s.push(current->left); //push left child
current = current->left;
flag.insert(current);
} //After iterate left tree, visit root
while(!s.empty() && s.top()->right == NULL){
current = s.top();
s.pop(); //pop root
ret.push_back(current->val); //visit root
}
if(s.empty()) break; //terminate when stack is empty
current = s.top();
s.pop(); //pop root
ret.push_back(current->val); //visit root //go to right child
s.push(current->right); //push right child
current = current->right;
} return ret;
} private:
vector<int> ret;
};
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