94. Binary Tree Inorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

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

递归方法：

/**
 * 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:
    void inOrderTraversal(vector& vi,TreeNode* root)
    {
        if(root->left) inOrderTraversal(vi,root->left);
        vi.push_back(root->val);
        if(root->right) inOrderTraversal(vi,root->right);
    }
    vector inorderTraversal(TreeNode* root) {
        vector vi;
        if(!root) return vi;
        inOrderTraversal(vi,root);
        
        return vi;
    }
};

使用栈

/**
 * 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 inorderTraversal(TreeNode* root) {
        vector vi;
        stack st;
        TreeNode* cur_node=root;
        while(!st.empty()||cur_node)
        {
            if(cur_node)
            {
                st.push(cur_node);
                cur_node=cur_node->left;
            }
            else
            {
                cur_node=st.top();
                st.pop();
                vi.push_back(cur_node->val);
                cur_node=cur_node->right;
            }
        }
        return vi;
    }
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
// LeetCode, Binary Tree Inorder Traversal
// Morris 中序遍历，时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
    public:
        vector inorderTraversal(TreeNode *root) {
            vector result;
            TreeNode *cur, *prev;
            cur = root;
            while (cur != nullptr) 
            {
                if (cur->left == nullptr) 
                {
                    result.push_back(cur->val);
                    prev = cur;
                    cur = cur->right;
                } 
                else 
                {
                    /* 查找前驱 */
                    TreeNode *node = cur->left;
                    while (node->right != nullptr && node->right != cur)
                        node = node->right;
                    if (node->right == nullptr) 
                    {   /* 还没线索化，则建立线索 */
                        node->right = cur;
                        /* prev = cur; 不能有这句， cur 还没有被访问 */
                        cur = cur->left;
                    } 
                    else 
                    { /* 已经线索化，则访问节点，并删除线索 */
                        result.push_back(cur->val);
                        node->right = nullptr;
                        prev = cur;
                        cur = cur->right;
                    }
                 }
            }
            return result;
    }
};