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/

5/8/2017

算法班

不要被难倒了，每个节点的left设为null就可以了。熟悉前序遍历

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

别人的答案

https://discuss.leetcode.com/topic/5783/accepted-simple-java-solution-iterative

第11，12行很好的方法，避免了prev

   public void flatten(TreeNode root) {
        if (root == null) return;
        Stack<TreeNode> stk = new Stack<TreeNode>();
        stk.push(root);
        while (!stk.isEmpty()){
            TreeNode curr = stk.pop();
            if (curr.right!=null)  
                 stk.push(curr.right);
            if (curr.left!=null)  
                 stk.push(curr.left);
            if (!stk.isEmpty()) 
                 curr.right = stk.peek();
            curr.left = null;  // dont forget this!! 
        }
    }

另外递归的方法

https://discuss.leetcode.com/topic/11444/my-short-post-order-traversal-java-solution-for-share

private TreeNode prev = null;

public void flatten(TreeNode root) {
    if (root == null)
        return;
    flatten(root.right);
    flatten(root.left);
    root.right = prev;
    root.left = null;
    prev = root;
}

更多讨论
https://discuss.leetcode.com/category/122/flatten-binary-tree-to-linked-list