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题目描述:(链接)
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?
解题思路：
迭代版:
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> result; 14 stack<TreeNode *> cache; 15 TreeNode *p = root; 16 17 while (!cache.empty() || p != nullptr) { 18 if (p != nullptr) { 19 cache.push(p); 20 p = p->left; 21 } else { 22 p = cache.top(); 23 cache.pop(); 24 result.push_back(p->val); 25 p = p->right; 26 } 27 } 28 29 return result; 30 } 31 };
递归版：
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 if (root != nullptr) { 14 inorderTraversal(root->left); 15 result.push_back(root->val); 16 inorderTraversal(root->right); 17 } 18 19 return result; 20 } 21 private: 22 vector<int> result; 23 };
[LeetCode]Binary Tree Inorder Traversal
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原文地址：http://www.cnblogs.com/skycore/p/4986158.html
