二叉树的遍历

raylan.chen大约 4 分钟

感谢 Carl 哥提供的中序遍历统一迭代法

感觉 Carl 哥提供的中序遍历统一迭代法（与普通的中序遍历迭代法相比，二叉树的迭代遍历 - 代码随想录）

跟二叉树的先序遍历（迭代遍历，二叉树的迭代遍历 - 代码随想录）较为一致，所以特地再去学了一遍

看完 Carl 哥的题解以及录友提供的 Java 题解，自己写了一遍

/**
 * 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) {
        if (root == null) return new ArrayList<Integer>();

        Deque<TreeNode> stack = new LinkedList<>();
        List<Integer> res = new ArrayList<>();
        stack.offerLast(root);
        while (!stack.isEmpty()) {
            //中序遍历，访问顺序 与 处理顺序 不同，需要特殊处理
            if (stack.peekLast() != null) {
                TreeNode tmp = stack.pollLast();

                if (tmp.right != null) stack.offerLast(tmp.right);
                //中间节点，null 为访问标记
                stack.offerLast(tmp);
                stack.offerLast(null);

                if (tmp.left != null) stack.offerLast(tmp.left);
            } else {    //该节点已经访问过了
                //弹空节点
                stack.pollLast();
                res.add(stack.pollLast().val);
            }
        }
        return res;
    }
}

想不到 LinkedList / Stack 还能存空节点，真不错，长见识了

（不过，不知道这会不会成为一个制造潜在 bug 的地方，比如处理空节点😂？刚开始就把 stack.peekLast() != null 写成了 stack != null，然后就判断空节点的右孩子是否为空了... ）

后序遍历

递归遍历

个人尝试（❌）

把递归跟迭代遍历搞混了，瞎写

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

    public List<Integer> postorderTraversal(TreeNode root) {
        if (root == null) return null;
        
        stack.offerLast(root);
        while (stack != null) {
            TreeNode tmp = stack.peek();
            
            postorderTraversal(stack.peek());
        }
    }
    public void postorder(TreeNode root) {
        if (root == null) return;

        while (stack != null) {
            TreeNode tmp = stack.peek();
            if (tmp.right != null) stack.offerLast(tmp.right);
            if (tmp.left != null) stack.offerLast(tmp.left);
            postorder(stack.peek());
            res.add(stack.pollLast());
        }
    }
}

随后，看了下 Carl 哥的讲解，自己再写了一遍

/**
 * 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> postorderTraversal(TreeNode root) {
        if (root == null) return new ArrayList<>();
        
        List<Integer> list = new ArrayList<>();
        postorder(root, list);
        return list;
    }

    public void postorder(TreeNode root, List<Integer> list) {
        if (root == null) return;
        postorder(root.left, list);
        postorder(root.right, list);
        list.add(root.val);
    }
}

迭代遍历

看了题解后，自己写了一遍

/**
 * 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> postorderTraversal(TreeNode root) {
        if (root == null) return new ArrayList<>();
        
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>(); 
        stack.offerLast(root);
        while (!stack.isEmpty()) {
            TreeNode tmp = stack.pollLast();
            res.add(tmp.val);
            if (tmp.left != null) stack.offerLast(tmp.left);
            if (tmp.right != null) stack.offerLast(tmp.right);
        }
        // 这里可以使用 Collections.reverse(res); 做题时不知道有这种用法，看了录友的题解才发现可以这样 
        for (int i = 0, j = res.size() - 1; i < j; i++, j--) {
            Integer tmp = res.get(i);
            res.set(i, res.get(j));
            res.set(j, tmp);
        }
        
        return res;
    }
}

优秀题解

二叉树的递归遍历 - 代码随想录

二叉树的迭代遍历 - 代码随想录

二叉树的遍历

二叉树的遍历

参考链接

中序遍历

迭代遍历（统一迭代）

后序遍历

递归遍历

迭代遍历

优秀题解