leetcode: Binary Tree Inorder Traversal

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

迭代实现：

/** * 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) {        TreeNode *p;        vector<int> res;        stack<TreeNode *> st;                if (root) {                        st.push(root);                        while (!st.empty()) {                                p = st.top();                if (p->left) {                    st.push(p->left);                    continue;                }                                while (!st.empty()) {                    p = st.top();                    st.pop();                    res.push_back(p->val);                                        if (p->right) {                        st.push(p->right);                        break;                    }                } //inter loop            } //outer loop        } //if block                return res;    }};

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> res;        vector<int> inorderTraversal(TreeNode *root) {                if (root) {            inorderTraversal(root->left);            res.push_back(root->val);            inorderTraversal(root->right);        }                return res;    }};