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/** * 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> vec; inOrder(root, vec); return vec; } /* void inOrder(TreeNode *root, vector<int> &path) { //递归写法 if (root) { inOrder(root->left, path); path.push_back(root->val); inOrder(root->right, path); } } */ //非递归写法 void inOrder(TreeNode *root, vector<int> &path) { stack<TreeNode *> TreeNodeStack; while (root != NULL || !TreeNodeStack.empty()) { while(root != NULL) { TreeNodeStack.push(root); root = root->left; } if (!TreeNodeStack.empty()) { root = TreeNodeStack.top(); TreeNodeStack.pop(); path.push_back(root->val); root = root->right; } } }};
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原文地址：http://www.cnblogs.com/gofighting/p/5411731.html
