114. Flatten Binary Tree to Linked List

Posted 2020-09-30

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.

树的题首先找容器(是全局变量还是输入值的递归 ), 再确定是分治(多后序) 还是递归(多前序),确定递归出口: 是否需要是找叶节点(left && right == null) 作正常出口(此时是输入值的递归见 257. Binary Tree Paths), 是否需要改变树的结构(此题借助pre改变左子树和上层节点作为右子树),  是否借助树的深度? 返回值是否为结果值, 或结果值作为全局变量, 返回中间值(见563 Binary Tree Tilt) 等等都是树的小知识加上容器, 加上先序? 后序? 中间值的做法.

public class Solution {
    TreeNode pre = null;
    public void flatten(TreeNode root) {
        if (root == null) {
            return;
        }
        helper(root);
    }
    private void helper(TreeNode root) {
        if (root == null) {
            return;
        }
        if (pre != null) {
            pre.left = null;
            pre.right = root;
        }
        pre = root;
        TreeNode right = root.right;
        
            helper(root.left);
        //}
        //if (right != null) {
            helper(right);
        //}
    }
}

　

if (root == null) {
            return;
        }  // 有了这句不用在判断左右是否为空再递归 : if (right != null) {

 

用先序遍历借助额外点即上层的点递的时候改变树的结构, 然后尾递归直接返回