标签：example linked public 二叉树 should
Total Accepted: 46999 Total Submissions: 163018My Submissions
Question Solution
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / 2 5 / \ 3 4 6
The flattened tree should look like:
1 2 3 4 5 6
分析：先序遍历，将二叉树转变为左支树
public class Solution {
TreeNode H;
void TravelTree(TreeNode x){
H.right=x;
H.left=null;
H=H.right;
TreeNode l=x.left;
TreeNode r=x.right;
x.left=null;
x.right=null;
if(l!=null)
TravelTree(l);
if(r!=null)
TravelTree(r);
}
public void flatten(TreeNode root) {
if(root!=null)
{
H=root;
TreeNode l=root.left;
TreeNode r=root.right;
root.left=null;
root.right=null;
if(l!=null)
TravelTree(l);
if(r!=null)
TravelTree(r);
}
/*
Stack<TreeNode> s1=new Stack<TreeNode>();
Stack<TreeNode> s2=new Stack<TreeNode>();
if(root!=null)
{
//TreeNode head=new TreeNode(0);
TreeNode h=root;
if(root.left!=null)
s1.push(root.left);
if(root.right!=null)
s1.push(root.right);
while(!s1.isEmpty())
{
TreeNode p=s1.peek();
s2.push(p);
s1.pop();
int mini=p.val;
TreeNode minitn=p;
while(!s1.isEmpty())
{
p=s1.peek();
s2.push(p);
s1.pop();
if(p.val<mini)
{
mini=p.val;
minitn=p;
}
}
while(!s2.isEmpty())
{
p=s2.peek();
if(p==minitn)
{
h.right=p;
h.left=null;
h=h.right;
s2.pop();
if(p.left!=null)
s1.push(p.left);
if(p.right!=null)
s1.push(p.right);
}
else
{
s2.pop();
s1.push(p);
}
}
}
}
*/
}
}
Leetcode#114Flatten Binary Tree to Linked List
标签：example linked public 二叉树 should
原文地址：http://7061299.blog.51cto.com/7051299/1653678
