[leetcode]Binary Tree Inorder Traversal

Morris遍历

就是利用节点的right指针

当前节点为curr

1. left为空，说明curr是子节点，所以输出，并且curr = curr->right

2. left不为空，tmp=curr->left 然后一直找tmp的右子树，知道tmp->right为空或者等于curr

    如果为空，说明还木有处理过，tmp->right = curr，然后curr = curr -> left

    不为空，说明前面已经处理过鸟，输出这个节点，然后设为空, curr = curr -> right（因为left遍历完了

http://www.cnblogs.com/AnnieKim/archive/2013/06/15/morristraversal.html

/**
 * 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* curr = root;
        vector<int> ans;
        if(root == nullptr) return ans;
        while(curr) {
            if(curr->left == nullptr) {
                ans.push_back(curr->val);
                curr = curr->right;
            } else {
                TreeNode* prev = curr->left;
                while(prev->right && prev->right!=curr) prev = prev->right;
                if(prev->right == nullptr) {
                    prev->right = curr;
                    curr = curr->left;
                } else {
                    ans.push_back(curr->val);
                    curr = curr->right;
                    prev->right = nullptr;
                }
            }
        }
        return ans;
    }
};