94. 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.

递归版本：

/** * 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> ret;        if(!root)            return ret;        if(root->left)            ret = inorderTraversal(root->left);        ret.push_back(root->val);        vector<int> right;        if(root->right)            right = inorderTraversal(root->right);        for(int i = 0;i<right.size();++i)            ret.push_back(right[i]);                    return ret;                                    }};

非递归版本：

根据中序遍历的顺序，对于任一结点，优先访问其左孩子，而左孩子结点又可以看做一根结点，然后继续访问其左孩子结点，直到遇到左孩子结点为空的结点才进行访问，然后按照相同的规则访问其 右子树。因此处理过程如下：

对任一结点p

1.若其左孩子不为空，则将p入栈并将p的左孩子置为当前的p，然后对当前结点p再进行相同的处理；

2.若其左孩子为空，则取栈顶元素并进行出栈操作，访问该栈顶结点，然后将栈顶结点的右孩子置为当前的p

3.直到p为null且栈为空则结束遍历

 * 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) {        stack<TreeNode * > s;    TreeNode *p = root;    vector<int> ret;        while(p != 0 || !s.empty())    {        while(p != 0)        {            s.push(p);            p = p -> left;        }        if(!s.empty())        {            p = s.top();            ret.push_back(p -> val);            s.pop();            p = p -> right;                    }    }        return ret;                }};