[leetcode]114. Flatten Binary Tree to Linked List@Java解题报告

https://leetcode.com/problems/flatten-binary-tree-to-linked-list/description/

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

click to show hints.

Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

本题有递归和循环解法

package go.jacob.day808;

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

/**
* 114. Flatten Binary Tree to Linked List
* @author Jacob
*
*/
public class Demo4 {
/*
* 解法一
* 递归解法
* 从尾到头逆序处理
*/
TreeNode pre=null;
public void flatten(TreeNode root) {
if(root==null)
return ;
flatten(root.right);
flatten(root.left);

root.left=null;
root.right=pre;
pre=root;

}
/*
* 解法二：循环求解
*/
public void flatten_1(TreeNode root) {
if(root==null)
return;
Stack<TreeNode> stack=new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node=stack.pop();
if(node.right!=null)
stack.push(node.right);
if(node.left!=null)
stack.push(node.left);
if(!stack.isEmpty())
node.right=stack.peek();
node.left=null;
}
}

/*
* 低效率解法
*/
public void flatten_2(TreeNode root) {
if(root==null)
return;
List<TreeNode> list=new ArrayList<TreeNode>();
Stack<TreeNode> stack=new Stack<TreeNode>();
stack.push(root);
TreeNode node=null;
while(!stack.isEmpty()||node!=null){
if(node==null){
node=stack.pop();
node=node.right;
}else{
stack.push(node);
list.add(node);
node=node.left;
}
}
TreeNode node1=null;
TreeNode node2=null;
for(int i=0;i<list.size()-1;i++){
node1=list.get(i);
node2=list.get(i+1);
node1.left=null;
node1.right=node2;
}
node2.left=null;node2.right=null;
}
}