【Leetcode】44. 二叉树的前序遍历

题目描述

题解

递归法

执行用时：0 ms, 在所有 Java 提交中击败了100.00%的用户

内存消耗：36.7 MB 在所有 Java 提交中击败了38.60%的用户

/**
 * 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 {
    List<Integer> res = new ArrayList<>();

    public List<Integer> preorderTraversal(TreeNode root) {
        preOrder(root);
        return res;
    }

    private void preOrder(TreeNode root) {
        if (root == null)
            return;
        res.add(root.val);
        preOrder(root.left);
        preOrder(root.right);
    }
}

递归都会写，主要还是考察遍历法 执行用时：0 ms, 在所有 Java 提交中击败了100.00%的用户

内存消耗：36.5 MB, 在所有 Java 提交中击败了84.27%的用户

/**
 * 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 {
    List<Integer> res = new ArrayList<>();

    public List<Integer> preorderTraversal(TreeNode root) {
        if (root == null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;

        while (!stack.isEmpty() || node != null) {
            while (node != null) {
                res.add(node.val);
                stack.push(node);
                node = node.left;
            }
            node = stack.pop();
            node = node.right;
        }
        return res;
    }
}