[LintCode] 二叉树的中序遍历

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

Another more sophisticated soltuion 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 {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in vector which contains node values.
     */
public:
    vector<int> inorderTraversal(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)) {
                    predecessor -> right = node;
                    node = node -> left;
                }
                else {
                    predecessor -> right = NULL;
                    nodes.push_back(node -> val);
                    node = node -> right;
                }
            }
            else {
                nodes.push_back(node -> val);
                node = node -> right;
            }
        }
        return nodes;
    }
};