LeetCode(面试题17.12)BiNode

虽然过了，但是还是不是很清楚，中序遍历，访问每个非空节点时，将左指针置空，pre记录每个父节点，父节点的right置为当前节点

/**
 * 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:
    TreeNode* convertBiNode(TreeNode* root) {
        TreeNode*p = root,*pre = NULL,*newroot;
        stack<TreeNode*>s;
        bool flag = true;
        while(p||!s.empty()){
            if(p){
                s.push(p);
                p = p->left;
            }else{
                p = s.top();
                if(flag&&p->left==NULL){
                    newroot = p;
                    flag = false;
                }
                p->left = NULL;
                if(pre)pre->right = p;
                cout<<p<<" "<<p->left<<" "<<p->right<<endl;
                s.pop();
                pre = p;
                p = p->right;
            }
        }
        return newroot;
    }
};

标签：pre,BiNode,面试题,right,TreeNode,17.12,NULL,节点,left
来源： https://www.cnblogs.com/stevenzrx/p/16286342.html