526互联

Binary Tree Inorder Traversal

发布时间 2023-09-02 14:31:28作者: 凌雨尘

Source

Given a binary tree, return the inorder traversal of its nodes' values.

Example
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

Challenge
Can you do it without recursion?

Note: Recursive solution is trivial, could you do it iteratively?

 

题解1 - 递归版

中序遍历的访问顺序为『先左再根后右』，递归版最好理解，递归调用时注意返回值和递归左右子树的顺序即可。

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        helper(root, result);
        return result;
    }

private:
    void helper(TreeNode *root, vector<int> &ret) {
        if (root != NULL) {
            helper(root->left, ret);
            ret.push_back(root->val);
            helper(root->right, ret);
        }
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        helper(root, result);
        return result;
    }

    private void helper(TreeNode root, List<Integer> ret) {
        if (root != null) {
            helper(root.left, ret);
            ret.add(root.val);
            helper(root.right, ret);
        }
    }
}

源码分析

Java 中 helper 的输入参数中 ret 不能和 inorderTraversal 中的 result 一样。

复杂度分析

树中每个节点都需要被访问常数次，时间复杂度近似为 O(n). 未使用额外辅助空间。
 

题解2 - 迭代版

使用辅助栈改写递归程序，中序遍历没有前序遍历好写，其中之一就在于入栈出栈的顺序和限制规则。我们采用「左根右」的访问顺序可知主要由如下四步构成。

  1. 首先需要一直对左子树迭代并将非空节点入栈
  2. 节点指针为空后不再入栈
  3. 当前节点为空时进行出栈操作，并访问栈顶节点
  4. 将当前指针p用其右子节点替代

步骤2,3,4对应「左根右」的遍历结构，只是此时的步骤2取的左值为空。

C++

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in vector which contains node values.
     */
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        stack<TreeNode *> s;

        while (!s.empty() || NULL != root) {
            if (root != NULL) {
                s.push(root);
                root = root->left;
            } else {
                root = s.top();
                s.pop();
                result.push_back(root->val);
                root = root->right;
            }
        }

        return result;
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        Deque<TreeNode> stack = new ArrayDeque<TreeNode>();

        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            if (curr != null) {
                stack.push(curr);
                curr = curr.left;
            } else {
                curr = stack.pop();
                result.add(curr.val);
                curr = curr.right;
            }
        }

        return result;
    }
}

源码分析

使用栈的思想模拟递归，注意迭代的演进和边界条件即可。Java 中新建变量 curr 而不是复用 root 观察下来有一点性能提升。

复杂度分析

最坏情况下栈保存所有节点，空间复杂度 O(n), 时间复杂度 O(n).