[LeetCode Hot 100] LeetCode94. 二叉树的中序遍历

题目描述

思路

熟练掌握迭代和递归的代码。

• 递归：额外写一个函数void inOrder(TreeNode node, List res)
• 迭代：令cur = root，一直往左子树找，找到最后一个左子节点，当cur为空，就开始处理栈顶元素(将栈顶元素加入结果集)，随后将cur设置为右子节点，继续执行以上操作。

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> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        inOrder(root, res);
        return res;
    }
    private void inOrder(TreeNode node, List<Integer> res) {
        if (node == null) return;
        inOrder(node.left, res);
        res.add(node.val);
        inOrder(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> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode cur = root;
        if (root == null) return res;

        while (cur != null || !stack.isEmpty()) {
            // 一直往左子树找,找到最后一个左子节点
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            // 当cur为空,就开始处理栈顶元素
            TreeNode node = stack.pop();
            res.add(node.val);
            // cur设置为当前节点的右子节点
            cur = node.right;
        }
        return res;
    }
}