Question ：

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?

confused what

"{1,#,2,3}"

means?



> read more on how binary tree is serialized on OJ.

Anwser 1 ：

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result(0);
        
        if (root == NULL)  return result;
        
        stack<TreeNode *> S;
        TreeNode* p = root;
        
        do
        {
            if (p != NULL) {
                S.push(p);
                p = p->left;
            } else {
                p = S.top();
                S.pop();
                
                result.push_back(p->val);
                p = p->right;
            }
        }while(!S.empty() || p != NULL);
        
        return result;
    }
};

Anwser 2 ：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    
    void dfs(TreeNode *p, vector<int> &result)
    {
        if(p->left != NULL)
            dfs(p->left, result);
            
        result.push_back(p->val);
        
        if(p->right!=NULL)
            dfs(p->right, result);
    }
    
    vector<int> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        TreeNode * head = root;
        vector<int> result;
        result.clear();
        
        if(head!=NULL) {
            dfs(head, result);
        }
            
        return result;
    }
};

Anwser 3 ：

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result;
        result.clear();
        stack<TreeNode *> stack_in;
        stack<TreeNode *> stack_out;
        
        if(root == NULL) return result;
        
        stack_in.push(root);
        while(!stack_in.empty())
        {
            TreeNode *node_in = stack_in.top();
            stack_in.pop();
            stack_out.push(node_in);
            if(node_in->right!=NULL && node_in->left!=NULL)
            {
                stack_in.push(node_in->right);
                stack_in.push(node_in->left);
            } else if(node_in->left!=NULL && node_in->right==NULL) {
                    stack_in.push(node_in->left);
            } else if(node_in->left==NULL && node_in->right!=NULL) {
                    stack_in.push(node_in->right);
                    result.push_back(node_in->val);
                    stack_out.pop();
            } else {
                result.push_back(node_in->val);
                stack_out.pop();
                while(!stack_out.empty())
                {
                    TreeNode * tmp = stack_out.top();
                    stack_out.pop();
                    result.push_back(tmp->val);
                    if(tmp->right!=NULL)
                        break;
                }
            }
        }
        return result;
    }
};

参考推荐：

Binary Tree Inorder Traversal