中序遍历二叉树

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

Example:

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

Output: [1,3,2]

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

方法一：递归遍历

代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> res;
    void inorder(TreeNode* root)
    {
        if(root)
        {
            inorder(root->left);
            res.push_back(root->val);
            inorder(root->right);
        }
    }
    vector<int> inorderTraversal(TreeNode* root) {
        res.clear();
        inorder(root);
        return res;
    }
};

方法二：迭代遍历，借助栈

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        TreeNode* curr = root;
        while(curr != NULL || !stk.empty())
        {
            while(curr != NULL)
            {
                stk.push(curr);
                curr = curr->left;
            }
            curr = stk.top(), stk.pop();
            res.push_back(curr->val);
            curr = curr->right;
        }
        return res;
    }
};

方法三：莫里斯法（线索二叉树）

1、如果当前结点的左孩子为空，则输出当前结点并将当前结点的右结点作为当前结点。
2、如果当前结点的左孩子不为空，则从当前结点的左子树找出当前结点的前驱节点：
如果前驱结点p的右孩子为空，则将p的右孩子设为当前结点；否则，输出当前结点，并将p的右孩子置为空，并将当前当前结点的右孩子置为当前结点
3、重复1 ，2两步直到当前结点为空

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        TreeNode* pre;
        while(root != NULL)
        {
            if(root->left == NULL)
            {
                res.push_back(root->val);
                root = root->right;
            }
            else
            {
                pre = root->left;
                while(pre->right != NULL && pre->right != root)
                {
                    pre = pre->right;
                }
                if(pre->right == NULL)
                {
                    pre->right = root;
                    root = root->left;
                }
                else
                {
                    pre->right = NULL;
                    res.push_back(root->val);
                    root = root->right;
                }
            }
        }
        return res;
    }
};