遍历二叉树

题目一：中序遍历
递归：

void inorder(Bitree *t){    if (t)    {        inorder(t->lchild);        printf("%c ", t->data);        inorder(t->rchild);    }}

非递归：

public class Solution {    /**     * @param root: The root of binary tree.     * @return: Inorder in ArrayList which contains node values.     */    public ArrayList<Integer> inorderTraversal(TreeNode root) {        Stack<TreeNode> stack = new Stack<TreeNode>();        ArrayList<Integer> result = new ArrayList<Integer>();        TreeNode curt = root;        while (curt != null || !stack.empty()) {            while (curt != null) {                stack.add(curt);                curt = curt.left;            }            curt = stack.peek();            stack.pop();            result.add(curt.val);            curt = curt.right;        }        return result;    }}

题目二：前序遍历
递归：

public class Solution {    public ArrayList<Integer> preorderTraversal(TreeNode root) {        ArrayList<Integer> result = new ArrayList<Integer>();        traverse(root, result);        return result;    }    // 把root为跟的preorder加入result里面    private void traverse(TreeNode root, ArrayList<Integer> result) {        if (root == null) {            return;        }        result.add(root.val);        traverse(root.left, result);        traverse(root.right, result);    }}

非递归：

public class Solution {    public List<Integer> preorderTraversal(TreeNode root) {        Stack<TreeNode> stack = new Stack<TreeNode>();        List<Integer> preorder = new ArrayList<Integer>();        if (root == null) {            return preorder;        }        stack.push(root);        while (!stack.empty()) {            TreeNode node = stack.pop();            preorder.add(node.val);            if (node.right != null) {                stack.push(node.right);            }            if (node.left != null) {                stack.push(node.left);            }        }        return preorder;    }}

分治：

public class Solution {    public ArrayList<Integer> preorderTraversal(TreeNode root) {        ArrayList<Integer> result = new ArrayList<Integer>();        // null or leaf        if (root == null) {            return result;        }        // Divide        ArrayList<Integer> left = preorderTraversal(root.left);        ArrayList<Integer> right = preorderTraversal(root.right);        // Conquer        result.add(root.val);        result.addAll(left);        result.addAll(right);        return result;    }}

题目三：后续遍历
递归：

public ArrayList<Integer> postorderTraversal(TreeNode root) {    ArrayList<Integer> result = new ArrayList<Integer>();    if (root == null) {        return result;    }    result.addAll(postorderTraversal(root.left));    result.addAll(postorderTraversal(root.right));    result.add(root.val);    return result;   }

非递归：

class Solution {public:    vector<int> postorderTraversal(TreeNode *root) {        vector<int> result;        stack<TreeNode *> myStack;        TreeNode *current = root, *lastVisited = NULL;        while (current != NULL || !myStack.empty()) {            while (current != NULL) {                myStack.push(current);                current = current->left;            }            current = myStack.top();             if (current->right == NULL || current->right == lastVisited) {                myStack.pop();                result.push_back(current->val);                lastVisited = current;                current = NULL;            } else {                current = current->right;            }        }        return result;    }};

题目四：层次遍历

public class Solution {    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {        ArrayList result = new ArrayList();        if (root == null) {            return result;        }        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.offer(root);        while (!queue.isEmpty()) {            ArrayList<Integer> level = new ArrayList<Integer>();            int size = queue.size();            for (int i = 0; i < size; i++) {                TreeNode head = queue.poll();                level.add(head.val);                if (head.left != null) {                    queue.offer(head.left);                }                if (head.right != null) {                    queue.offer(head.right);                }            }            result.add(level);        }        return result;    }}

题目五：二叉树迭代器

class BSTIterator {public:    stack<TreeNode *> myStack;    TreeNode *current;    BSTIterator(TreeNode *root) {        while (!myStack.empty()) {            myStack.pop();        }        current = root;    }    /** @return whether we have a next smallest number */    bool hasNext() {        return (current != NULL || !myStack.empty());    }    /** @return the next smallest number */    int next() {        while (current != NULL) {            myStack.push(current);            current = current->left;        }        current = myStack.top(); myStack.pop();        int val = current->val;        current = current->right;        return val;    }};