114. Flatten Binary Tree to Linked List【Medium】【将给定的二叉树转化为“只有右孩子节点”的链表（树）】...

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1/ \2   5/ \   \
3   4   6

The flattened tree should look like:

1\2\3\4\5\6

Accepted

218,918

Submissions

533,947

【解析】由上图可知，如果右子树不为空，则右子树最后肯定为左子树最有一个靠右的孩子节点的右子树，而左子树最后成为整棵树的右子树。这样，首先判断左子树是否为空，不为空就寻找到树根的左孩子节点，然后寻找该节点是否有右孩子，如果有继续寻找，直到找到属于叶子节点的右孩子，此时，该节点的右子树“指向”当前树的右子树，并将当前左子树变为树根的右孩子，将整棵树左孩子置为空。最后，根节点“指向”根节点的右孩子，继续上述操作，直到整棵树遍历完即得到结果。

1.首先，将根节点的右子树，连接到根节点的左子树的最右节点，如下图所示：

2.然后将左子树移动到右边，如下图所示：

3.之后，将根节点的右节点作为当前的根节点，循环进行。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public void flatten(TreeNode root) {while(root != null) {if(root.left != null) {TreeNode pre = root.left;while(pre.right != null) {pre = pre.right;}pre.right = root.right;root.right = root.left;root.left = null;}root = root.right;}}
}

转载于:https://www.cnblogs.com/Roni-i/p/10434572.html

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