114. Flatten Binary Tree to Linked List

题目：

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.

链接： http://leetcode.com/problems/flatten-binary-tree-to-linked-list/

题解：

使用栈来辅助，很像先序遍历。也可以用recursive和Morris-travel。

Time Complexity - O(n)， Space Complexity - O(n)。

public class Solution {
    public void flatten(TreeNode root) {
        if(root == null)
            return;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        
        while(node != null || !stack.isEmpty()){
            if(node.right != null){
                stack.push(node.right);
            }
            if(node.left != null){
                node.right = node.left;
                node.left = null;
            } else {
                if(!stack.isEmpty())
                    node.right = stack.pop();
            }
            node = node.right;
        }
    }
}

 

二刷：

先建立一个Stack<TreeNode>()， 再取得一个root的reference node。在root != null并且!stack.isEmpty()的情况下进行遍历。

假如右子树非空，则入栈。  

假如左子树非空，则将左子树放到右子树处，置空左子树。 否则左子树为空时，假如stack非空，则我们设置node.right = stack.pop();

将node移动一位 - node = node.right。

Discuss区有很多好解法。下次一定要补上recursive的。

Java:

Time Complexity - O(n)， Space Complexity - O(n)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void flatten(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while (node != null) {
            if (node.right != null) stack.push(node.right);
            if (node.left != null) {
                node.right = node.left;
                node.left = null;
            } else if (!stack.isEmpty()){
                node.right = stack.pop();
            }
            node = node.right;
        }
    }
}

 

Reference:

https://leetcode.com/discuss/30719/my-short-post-order-traversal-java-solution-for-share

https://leetcode.com/discuss/17944/accepted-simple-java-solution-iterative

https://leetcode.com/discuss/27643/straightforward-java-solution

https://leetcode.com/discuss/13054/share-my-simple-non-recursive-solution-o-1-space-complexity