LeetCode 94. 二叉树的中序遍历

迭代

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> inorderTraversal(TreeNode* root) {
13        vector<int> ans;
14         if(!root) return ans;
15         stack<TreeNode*> nodeStack;
16         while(root) 
17         {
18             nodeStack.push(root);
19             root = root->left;
20         }
21         while(!nodeStack.empty()) {
22             TreeNode* node = nodeStack.top();
23             ans.push_back(node->val);
24             nodeStack.pop();
25             if(node = node->right) {
26                 while(node) {
27                     nodeStack.push(node);
28                     node = node->left;
29                 } 
30             }
31         }
32 
33         return ans;
34     }
35 };

优化：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> st; 
        while (root!=NULL || !st.empty()) {
            while (root != NULL) {
                st.push(root);
                root = root->left;
            }
            root = st.top();
            st.pop();
            res.push_back(root->val);
            root = root->right;
        }
        return res;
    }
};

递归：

 1 class Solution {
 2 public:
 3      void process(TreeNode* root,vector<int> &ans)
 4     {
 5       
 6         if(root->left) process(root->left,ans);
 7           ans.push_back(root->val);
 8         if(root->right) process(root->right,ans);
 9     }
10 
11     vector<int> inorderTraversal(TreeNode* root) {
12         vector<int> ans;
13         if(root)
14             process(root,ans);
15         return ans;
16     }
17 };