06
08/2015
[LeetCode] Binary Tree Inorder Traversal
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].
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
解题思路：
中序遍历，递归算法没有啥问题，先左子树，然后根节点，然后右子树。
/**
* 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) {
vector<int> result;
helper(root, result);
return result;
}
void helper(TreeNode* root, vector<int>& result){
if(root == NULL) {
return;
}
helper(root->left, result);
result.push_back(root->val);
helper(root->right, result);
}
};迭代版本：
/**
* 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) {
vector<int> result;
stack<TreeNode*> s;
TreeNode* node = root;
while(!s.empty() || node!=NULL){
while(node!=NULL){
s.push(node);
node = node->left;
}
node = s.top();
s.pop();
result.push_back(node->val);
node = node->right;
}
return result;
}
};
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