LintCode 二叉树的中序遍历

给出一棵二叉树,返回其中序遍历

样例

给出二叉树 {1,#,2,3},

   1
    
     2
    /
   3

返回 [1,3,2].

挑战 

你能使用非递归算法来实现么?

分析:同前序遍历。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in vector which contains node values.
     */
public:
    vector<int> inorderTraversal(TreeNode *root) {
        // write your code here
      TreeNode *curr=root;
      stack<TreeNode *> mystack;
      vector<int> res;
      while(!mystack.empty()||curr!=NULL)
      {
          while(curr!=NULL)
          {
              
              mystack.push(curr);
              curr=curr->left;
          }
          if(!mystack.empty())
          {
              curr=mystack.top();
              res.push_back(curr->val);
              mystack.pop();
              curr=curr->right;
          }
      }
      return res;
    }
};


用数组指针

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in vector which contains node values.
     */
public:
    vector<int> inorderTraversal(TreeNode *root) {
        // write your code here
         TreeNode *curr=root;
         TreeNode *mystack[1000];
         int top=0;
         vector<int> res;
         while(top!=0||curr!=NULL)
         {
             while(curr!=NULL)
            { 
             
             mystack[top++]=curr;
             curr=curr->left;
            }
         
         if(top>0)
         {
             top--;
             curr=mystack[top]; 
             res.push_back(curr->val);
             curr=curr->right;
             
         }
         }
         return res;
    }
};