114. Flatten Binary Tree to Linked List (Medium)

453. Flatten Binary Tree to Linked List (Easy)

解法1: 用stack.

class Solution {
public:
    void flatten(TreeNode *root) {
    	if(!root) return;
    	TreeNode *pnode = NULL;
    	stack<TreeNode*> s;
    	s.push(root);
    	while(!s.empty()){
    		root = s.top(); s.pop();
    		if(pnode){
    			pnode->left = NULL;
    			pnode->right = root;
    		}
    		if(root->right) s.push(root->right);
    		if(root->left) s.push(root->left);
    		pnode = root;
    	}
    	pnode->left = pnode->right = NULL;
    }
};

解法2: 递归

class Solution {
private:
    void _flatten(TreeNode *root, TreeNode **ph, TreeNode **pt) {
        if (!root) {
            *ph = *pt = NULL;
            return;
        }
        TreeNode *h1, *t1, *h2, *t2;
        _flatten(root->left, &h1, &t1);
        _flatten(root->right, &h2, &t2);
        root->left = NULL;
        *ph = *pt = root;
        if (h1) {
            root->right = h1;
            *pt = t1;
            if (h2) {
                t1->right = h2;
                *pt = t2;
            }
        } else if (h2) {
            root->right = h2;
            *pt = t2;
        }
    }
public:
    void flatten(TreeNode *root) {
        TreeNode *h, *t;
        _flatten(root, &h, &t);
    }
};

ph其实一定指向root, 所以可以省略. 简化为一下代码:

class Solution {
private:
    void _flatten(TreeNode *root, TreeNode **ptail) {
        if (!root) {
            *ptail = NULL;
            return;
        }
        TreeNode *t1, *t2;
        _flatten(root->left, &t1);
        _flatten(root->right, &t2);
        if (t1) {
            t1->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
        *ptail = t2 ? t2 : (t1 ? t1 : root);
    }
public:
    void flatten(TreeNode *root) {
        TreeNode *t;
        _flatten(root, &t);
    }
};

解法3: @Netario36

假设flatten函数已经可以以前序遍历完成flatten的任务, 且flatten(root->left)和flatten(root->right)已完成, 那么接下来要做的操作就是:

记flatten(root->left)后该链表的最后一个节点是tail.

那么

tail->right = root->right;
root->right = tail;
root->left = NULL;

就可以把root节点, root->left的flatten子树, 和root->right的flatten子树串起来.

这代码看起来简洁, 但是把左子树处理完并插入到root和root->right之间后, 要处理插入前的root->right子树需要递归N次, 其中N是root->left子树节点数. 也就是说, 有很多不必要的递归.

//@Netario36
class Solution {
private:
    TreeNode *tail;
public:
    void flatten(TreeNode *root) {
        if (!root) return;
        if (root->left) {
            flatten(root->left);
            tail->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
        tail = root;
        flatten(root->right);
    }
};

基于@Netario36的版本我写了一个前序遍历结构更明显, 没有无用的递归的版本.

class Solution {
private:
    TreeNode *tail;
public:
    void flatten(TreeNode *root) {
        if (!root) return;
        tail = root;
        if (root->left) {
            flatten(root->left);
            tail->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
        flatten(tail->right);
    }
};

解法4: @versavitality

参数rightRoot是root的右兄弟节点, 返回值为root子树与rightRoot子树flatten之后的根节点.

TreeNode *right = _flatten(root->right, rightRoot);将root->right子树与rightRoot子树flatten到一起.

root->right = _flatten(root->left, right);将root->left子树与上一步的结果flatten到一起.

// @versavitality
class Solution {
private:
    TreeNode* _flatten(TreeNode *root, TreeNode *rightRoot) {
        if (!root) return rightRoot;
        TreeNode *right = _flatten(root->right, rightRoot);
        root->right = _flatten(root->left, right);
        root->left = NULL;
        return root;
    }
public:
    void flatten(TreeNode *root) {
        _flatten(root, NULL);
    }
};

蛋疼的博客园, 有@字符会翻译成mailto链接…又不知道怎么取消, 只能用``包起来了.