[LintCode] 二叉树的前序遍历

The recursive solution is trivial and I omit it here.

Iterative Solution using Stack (O(n) time and O(n) space):

/**
* Definition of TreeNode:
* class TreeNode {
* public:
*     int val;
*     TreeNode *left, *right;
*     TreeNode(int val) {
*         this->val = val;
*         this->left = this->right = NULL;
*     }
* }
*/

class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: Preorder in vector which contains node values.
*/
vector<int> preorderTraversal(TreeNode *root) {
// write your code here
vector<int> nodes;
TreeNode* node = root;
stack<TreeNode*> right;
while (node || !right.empty()) {
if (node) {
nodes.push_back(node -> val);
if (node -> right)
right.push(node -> right);
node = node -> left;
}
else {
node = right.top();
right.pop();
}
}
return nodes;
}
};

Another more sophisticated solution using Morris Traversal (O(n) time and O(1) space):

/**
* Definition of TreeNode:
* class TreeNode {
* public:
*     int val;
*     TreeNode *left, *right;
*     TreeNode(int val) {
*         this->val = val;
*         this->left = this->right = NULL;
*     }
* }
*/

class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: Preorder in vector which contains node values.
*/
vector<int> preorderTraversal(TreeNode *root) {
// write your code here
vector<int> nodes;
TreeNode* node = root;
while (node) {
if (node -> left) {
TreeNode* predecessor = node -> left;
while (predecessor -> right && predecessor -> right != node)
predecessor = predecessor -> right;
if (!(predecessor -> right)) {
nodes.push_back(node -> val);
predecessor -> right = node;
node = node -> left;
}
else {
predecessor -> right = NULL;
node = node -> right;
}
}
else {
nodes.push_back(node -> val);
node = node -> right;
}
}
return nodes;
}
};