分解让复杂问题简单化：二叉搜索树与双向链表

/**
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;

public TreeNode(int val) {
this.val = val;

}

}
*/
public class Solution {
public TreeNode Convert(TreeNode pRootOfTree) {
if (pRootOfTree == null) {
return null;
}
if (pRootOfTree.left == null && pRootOfTree.right == null) {
return pRootOfTree;
}
// 1.将左子树构造成双链表，并返回链表头节点
TreeNode left = Convert(pRootOfTree.left);
TreeNode p = left;
// 2.定位至左子树双链表最后一个节点
while (p != null && p.right != null) {
p = p.right;
}
// 3.如果左子树链表不为空的话，将当前root追加到左子树链表
if (left != null) {
p.right = pRootOfTree;
pRootOfTree.left = p;
}
// 4.将右子树构造成双链表，并返回链表头节点
TreeNode right = Convert(pRootOfTree.right);
// 5.如果右子树链表不为空的话，将该链表追加到root节点之后
if (right != null) {
right.left = pRootOfTree;
pRootOfTree.right = right;
}
return left != null ? left : pRootOfTree;

}
}