[leetcode] 树的前序、中序的iterative traversal

Posted by: lexigrey on: August 4, 2013

• In: classic | leetcode
• 1 Comment

经典中的经典。

前序 pre order （中左右）：

1. push root
2. pop，若有right, push right，若有left push left
3. 这样继续一边pop一边push。直到stack为空。

public void preOrder(TreeNode root) {
  if (root == null)
    return;
  Stack<TreeNode> s = new Stack<TreeNode>();
  s.push(root);
  while (!s.isEmpty()) {
    TreeNode top = s.pop();
    System.out.print(top.val + ", ");
    if (top.right != null)
      s.push(top.right);
    if (top.left != null)
      s.push(top.left);
  }
}

中序 in order （左中右）：

1. push root左支到死
2. pop， 若pop的有right，则把right当root对待push它的左支到死。
3. 这样继续一边pop一边push。直到stack为空。

public void inOrder(TreeNode root) {
  if (root == null)
    return;
  Stack<TreeNode> s = new Stack<TreeNode>();
  pushAllTheyWayAtLeft(s, root);
  while (!s.isEmpty()) {
    TreeNode top = s.pop();
    System.out.print(top.val + ", ");
    if (top.right != null)
      pushAllTheyWayAtLeft(s, top.right);
  }
}
private void pushAllTheyWayAtLeft(Stack<TreeNode> s, TreeNode root) {
  while (root != null) {
    s.push(root);
    root = root.left;
  }
}

another way: keep a curr pointer, when curr is null means reached the left most, then start popping

    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode curr = root;
        while (!stack.isEmpty() || curr != null) {
            if (curr != null) {
                stack.push(curr);
                curr = curr.left;
            } else {
                curr = stack.pop();
                result.add(curr.val);
                curr = curr.right;
            }
        }
        return result;
    }

Tags: converToIteration, tree