原题:
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / \ 2 5 / \ \ 3 4 6The flattened tree should look like:
1 \ 2 \ 3 \ 4 \ 5 \ 6
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
思路:第一反应是深度优先遍历,依次把节点放到树叶子的右子树位置即可。但是看到hint里提到的是先序遍历,就百度了下差别,现在概念更加清晰啦~
【先序,后序,中序针对二叉树。深度、广度针对普通树。】
代码如下:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void flatten(TreeNode *root) {
if (!root) return;
TreeNode *tmp;
queue<TreeNode *> res;
//depth first search
stack<TreeNode*> tree;
tree.push(root);
while(!tree.empty()){
tmp = tree.top();
tree.pop();
res.push(tmp);
if(tmp->right!=NULL) tree.push(tmp->right);
if(tmp->left!=NULL) tree.push(tmp->left);
}
root = res.front();
res.pop();
tmp = root;
while(!res.empty()){
tmp->right = res.front();
tmp->left = NULL;
tmp = tmp->right;
res.pop();
}
}
};