二叉樹的中序遍歷（遞歸+迭代）

題目

給定一個二叉樹，返回它的 中序 遍歷。

示例:

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

輸出: [1,3,2]

進階: 遞歸算法很簡單，你可以通過迭代算法完成嗎？

遞歸

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    List<Integer> ans;
    
    public List<Integer> inorderTraversal(TreeNode root) {
        ans = new ArrayList<>();
        dfs(root);
        return ans;
    }
    
    public void dfs(TreeNode root) {
        if (root == null) return ;
        dfs(root.left);
        ans.add(root.val);
        dfs(root.right);
    }
}

迭代1

自己寫的迭代，就是仿造遞歸的順序寫的：

/**
 * 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> inorderTraversal(TreeNode root) {
        
        List<Integer> ans = new ArrayList<>();
        LinkedList<TreeNode> stack = new LinkedList<>();
        if (root == null) return ans;
        
        // 把最左邊的節點入棧，但並不訪問
        TreeNode p = root;
        while (p != null) {
            stack.add(p);
            p = p.left;
        }     
        
        // 訪問並出棧，若有右孩子，將右孩子最左邊節點入棧
        while (!stack.isEmpty()) {
            p = stack.pollLast();
            ans.add(p.val);
            p = p.right;
            while (p != null) {
                stack.add(p);
                p = p.left;
            }
        }
        
        return ans;
    }
}

迭代2

官方題解的迭代，缺點是用了Stack，這個類是不推薦使用的，優點是比我寫的優雅T^T

public class Solution {
    public List < Integer > inorderTraversal(TreeNode root) {
    
        List < Integer > res = new ArrayList<>();
        Stack < TreeNode > stack = new Stack<>();
        TreeNode curr = root;
        
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            res.add(curr.val);
            curr = curr.right;
        }
        
        return res;
    }
}