Binary Tree Traversal

Preorder:

Approach #1: Recurisive.

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        helper(root, ans);
        return ans;
    }
    
private:
    void helper(TreeNode* root, vector<int>& ans) {
        if (root == NULL) return ;
        ans.push_back(root->val);
        helper(root->left, ans);
        helper(root->right, ans);  
    }
};

Approach #2: Iteratively.[Java]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<Integer>();
        Stack<TreeNode> todo = new Stack<TreeNode>();
        TreeNode cur = root;
        while (cur != null) {
            res.add(cur.val);
            if (cur.right != null) {
                todo.push(cur.right);
            }
            cur = cur.left;
            if (cur == null && !todo.isEmpty()) {
                cur = todo.pop();
            }
        }
        return res;
    }
}

Approach #3: Morris Traversal.[C++]

/**
 * 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> preorderTraversal(TreeNode* root) {
        TreeNode* cur = root;
        vector<int> nodes;
        while (cur) {
            if (cur->left) {
                TreeNode* pre = cur->left;
                while (pre->right && (pre->right != cur)) {
                    pre = pre->right;
                }
                if (!(pre->right)) {
                    nodes.push_back(cur->val);
                    pre->right = cur;
                    cur = cur->left;
                } else {
                    pre->right = NULL;
                    cur = cur->right;
                }
            } else {
                nodes.push_back(cur->val);
                cur = cur->right;
            }
        }
        return nodes;
    }
};

Using Morris Traversal can don't use recurisive and stack and space complex is O(1).

---

inorder

Approach #1: Java.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res =  new ArrayList< > ();
        Stack<TreeNode> stack = new Stack< > ();
        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            res.add(curr.val);
            curr = curr.right;
        }
        return res;
    }
}

　　

---

postorder

Approach #2: C++.

/**
 * 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> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> todo;
        vector<int> res;
        TreeNode* node = root;
        TreeNode* last = NULL;
        while (node || !todo.empty()) {
            if (node) {
                todo.push(node);
                node = node->left;
            } else {
                TreeNode* top = todo.top();
                if (top->right && top->right != last) {
                    node = top->right; 
                } else {
                    res.push_back(top->val);
                    last = top;
                    todo.pop();
                }
            }
        }
        return res;
    }
};

　　

---

Binary Tree Level Order Traversal

Approach #1: C++. [Using queue]

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        if (root == NULL) return ans;
        q.push(root);
        while (!q.empty()) {
            int size = q.size();
            vector<int> dummy;
            for (int i = 0; i < size; ++i) {
                TreeNode* node = q.front();
                if (node->left != NULL) q.push(node->left);
                if (node->right != NULL) q.push(node->right);
                dummy.push_back(node->val);
                q.pop();
            }
            ans.push_back(dummy);
        }
        
        return ans;
    }
};

　　

Approach #2: Java. [Using recursive.] 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        levelHelper(res, root, 0);
        return res;
    }
    
    public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
        if (root == null) return;
        if (height >= res.size()) {
            res.add(new LinkedList<Integer>());
        }
        res.get(height).add(root.val);
        levelHelper(res, root.left, height + 1);
        levelHelper(res, root.right, height + 1);
    }
}