[LeetCode Hot 100] LeetCode144. 二叉树的前序遍历

题目描述

思路

熟练掌握迭代和递归的代码。

• 递归代码：额外写一个函数void preOrder(TreeNode node, List res)
• 迭代代码：会用到数据结构——栈。先入栈当前节点的右子节点，再入栈左子节点。

方法一：递归

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

方法二：迭代

/**
 * 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> res = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        
        if (root != null) stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            res.add(node.val);
			// 先入栈当前节点的右子节点，再入栈左子节点。
            if (node.right != null) stack.push(node.right);
            if (node.left != null) stack.push(node.left);
        }

        return res;
    }
}