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leetcode 144. binary tree preorder traversal

admin 11月 13, 2020 0

144. Binary Tree Preorder Traversal

Difficulty:: Medium

Given a binary tree, return the preorder traversal of its nodes’ values.

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

Solution

Language: Java


 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        traverse(root, result);
        return result;
    }
    
    private void traverse(TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
    
    // 非递归写法1
    private void traverseIterative(TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        Deque<TreeNode> s = new ArrayDeque<>();
        s.push(root);
        while (!s.isEmpty()) {
            TreeNode tmp = s.pop();
            result.add(tmp.val);
            if (tmp.right != null) {
                s.push(tmp.right);
            }
            if (tmp.left != null) {
                s.push(tmp.left);
            }
        }
    }
    
    // 非递归写法2
    private void traverseIterative(TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        Deque<TreeNode> s = new ArrayDeque<>();
        TreeNode cur = root;
        while (!s.isEmpty() || cur != null) {
            if (cur != null) {
                s.push(cur);
                result.add(cur.val);
                cur = cur.left;
            } else {
                cur = s.pop();
                cur = cur.right;
            }
        }
    }
}