[LeetCode] 94 - 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?
/**
* 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:
#if 0
void doInorder(TreeNode *node, vector<int> &res) {
if (node->left) doInorder(node->left, res);
res.push_back(node->val);
if (node->right) doInorder(node->right, res);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
if (!root) return res;
doInorder(root, res);
return res;
}
#else
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
if (!root) return res;
stack<TreeNode* > treeStack;
do {
while (root) {
treeStack.push(root);
root = root->left;
}
if (!treeStack.empty()) {
root = treeStack.top();
treeStack.pop();
res.push_back(root->val);
root = root->right;
}
} while(root || !treeStack.empty());
return res;
}
#endif
};
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