Leetcode: 94. Binary Tree Inorder Traversal

Description

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

Example

Given binary tree [1,null,2,3],

1
 \
  2
 /
3

return [1,3,2].

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

思路

• 使用一個棧來保存節點，而後使用一個map來標記已經遍歷過的節點

代碼

/**
 * 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> res;
        if(!root) return res;
        
        stack<TreeNode*> stk;
        stk.push(root);
        TreeNode *tmp = NULL;
        unordered_map<TreeNode*, bool> hash;
        hash[root] = true;
        while(!stk.empty()){
            tmp = stk.top();
           
            while(tmp->left && hash.count(tmp->left) == 0){
                stk.push(tmp->left);
                hash[tmp->left] = true;
                tmp = tmp->left;
            }
            
            tmp = stk.top();
            stk.pop();
            
            if(tmp->right)
                stk.push(tmp->right);
            res.push_back(tmp->val);
        }
        
        return res;
    }
};