GuilinDev

Lc0144

05 August 2008

144 Binary Tree Preorder Traversal

原题概述

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

Example:

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

Output: [1,2,3]

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

题意和分析

前序遍历，根左右。

代码

递归

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/**
 * 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> result = new ArrayList<>();
        preorderTraversalHelper(root, result);
        return result;
    }
    private void preorderTraversalHelper(TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        result.add(root.val);
        preorderTraversalHelper(root.left, result);
        preorderTraversalHelper(root.right, result);
    }
}

迭代

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        
        while (!stack.isEmpty()) {
            root = stack.pop();
            if (root == null) {
                continue;
            }
            stack.push(root.right);
            stack.push(root.left);
            if (root.left == null && root.right == null) {
                result.add(root.val);
            } else {
                stack.push(new TreeNode(root.val)); // 这里不是直接加入root，而是新建一个node
            }            
        }
        return result;
    }
}
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