leetcode: Binary Tree Inorder Traversal

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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].

递归版：

/** * 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;        if(!root)            return res;        core( root, res);        return res;    }    void core( TreeNode *root, vector< int> &res){        if( root == NULL)            return;        core( root->left, res);        res.push_back(root->val);        core( root->right, res);    }};

迭代版，用stack模拟递归

/** * 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;        if( !root)            return res;        stack< TreeNode *> stk;        stk.push(root);        while( !stk.empty()){            TreeNode *p = stk.top();            stk.pop();            if( p->left == NULL && p->right == NULL){                res.push_back(p->val);            }            else{                if( p->right)                    stk.push(p->right);                stk.push(p);                if( p->left)                    stk.push(p->left);                p->left = p->right = NULL;//这句不能少，因为之前已经将p的左儿子，有儿子放入栈里，如果这里不置为NULL，那么会无限循环，不断把p左右儿子放入栈中            }        }        return res;    }};

标准非递归：

/** * 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;        if( !root)            return res;        stack< TreeNode *> stk;        TreeNode *p = root;        while(true){            while(p){                stk.push(p);                p = p->left;            }            if( !stk.empty()){                p = stk.top();                stk.pop();                res.push_back(p->val);                p = p->right;            }            else                break;        }        return res;    }};

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