【二叉树遍历】中序

记录下可以用的中序遍历方法

方法1：

vector<int> InOrderTraverse(TreeNode * root) //中序遍历
    {
        vector<int> ans;
        TreeNode* p;
        vector<TreeNode *> Stack;
        Stack.push_back(root);
        while(!Stack.empty())
        {
            while((p = Stack.back()) != NULL) 
                Stack.push_back(p->left);
            Stack.pop_back();
            if(!Stack.empty())
            {
                p = Stack.back(); 
                Stack.pop_back();
                ans.push_back(p->val);
                Stack.push_back(p->right);
            }

        }
        return ans;
    }

方法2：

vector<int> inorderTraversal(TreeNode *root) {
        vector<int> vector;
        stack<TreeNode *> stack;
        TreeNode *pCurrent = root;

        while(!stack.empty() || pCurrent)
        {
            if(pCurrent)
            {
                stack.push(pCurrent);
                pCurrent = pCurrent->left;
            }
            else
            {
                TreeNode *pNode = stack.top();
                vector.push_back(pNode->val);
                stack.pop();
                pCurrent = pNode->right;
            }
        }
        return vector;
    }

我按照自己的思路写的，调了很多遍。关键是右子树弹出时候的逻辑混乱

vector<int> inorderTraversal(TreeNode *root) {
        vector<int> ans;
        vector<TreeNode *> v;
        TreeNode * p;
        if(root == NULL) return ans;
        v.push_back(root);

        while(!v.empty())
        {
            p = v.back();
            while(p->left != NULL) //找到最左边的
            {
                v.push_back(p->left);
                p = p->left;
            }

            ans.push_back(p->val); //压入数据
            v.pop_back();

            while(p->right == NULL) //右子树处理 如果当前的右子树是空的 那么需要弹出v中栈尾的数值
            {
                if(!v.empty())
                {
                    p = v.back();
                    ans.push_back(p->val);
                    v.pop_back();
                }
                else
                {
                    break;
                }
            }
            if(p->right != NULL)
            {
                v.push_back(p->right);
            }
        }

        return ans;
    }

还有不需要栈的方法，利用了线索二叉树的概念，空间复杂度O(1)

http://www.cnblogs.com/AnnieKim/archive/2013/06/15/morristraversal.html