【LeetCode】94. Binary Tree Inorder Traversal (3 solutions)

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?

解法一：递归

/**
 * Definition for binary tree
 * 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> ret;
        if(!root)
            return ret;
        inorder(root, ret);
        return ret;
    }
    void inorder(TreeNode* root, vector<int>& ret)
    {
        if(root->left)
            inorder(root->left, ret);
        ret.push_back(root->val);
        if(root->right)
            inorder(root->right, ret);
    }
};

解法二：非递归

使用map记录是否访问过，使用栈记录访问路径，访问过就出栈。

遵循左根右原则。

/**
 * Definition for binary tree
 * 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> ret;
        if(!root)
            return ret;
            
        stack<TreeNode*> stk;
        unordered_map<TreeNode*, bool> m;
        stk.push(root);
        m[root] = true;
        while(!stk.empty())
        {
            TreeNode* top = stk.top();
            if(top->left)
            {
                if(m[top->left] == false)
                {
                    stk.push(top->left);
                    m[top->left] = true;
                    continue;
                }
            }
            ret.push_back(top->val);
            stk.pop();
            if(top->right)
            {
                if(m[top->right] == false)
                {
                    stk.push(top->right);
                    m[top->right] = true;
                }
            }
        }
        return ret;
    }
};

解法三：非递归，不需要除栈之外的空间

每次新加入节点时，将左子节点（比当前节点小）全部进栈，这样在出栈的时候就不需要再去访问左子节点，

只需要访问右子节点（比当前节点大）

/**
 * 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> ret;
        if(root == NULL)
            return ret;
        stack<TreeNode*> stk;
        stk.push(root);
        TreeNode* cur = root;
        while(cur->left)
        {
            stk.push(cur->left);
            cur = cur->left;
        }
        while(!stk.empty())
        {
            TreeNode* top = stk.top();
            stk.pop();
            ret.push_back(top->val);
            if(top->right)
            {
                TreeNode* cur = top->right;
                stk.push(cur);
                while(cur->left)
                {
                    stk.push(cur->left);
                    cur = cur->left;
                }
            }
        }
        return ret;
    }
};