[leetcode][tree] Binary Tree Inorder Traversal

题目：

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?

递归实现：

/** * Definition for binary tree * 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;        inorderTraversalCore(root, res);        return res;    }private:    void inorderTraversalCore(TreeNode *root, vector<int> &res){        if(NULL == root) return;        inorderTraversalCore(root->left, res);        res.push_back(root->val);        inorderTraversalCore(root->right, res);    }};

非递归实现（栈+循环）

/** * 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 (NULL == root) return res;stack<TreeNode *> sta;TreeNode *p = root;//用于遍历，指向当前节点while (p || !sta.empty()){//注意循环结束条件while (p){//如果有左孩子，一直入栈（找到最左节点）sta.push(p);p = p->left;}p = sta.top();//访问栈顶元素sta.pop();res.push_back(p->val);p = p->right;//考察栈顶元素的右孩子}return res;}};