[Lintcode]Inorder Successor in Binary Search Tree(DFS)

题意

略

分析

1.首先要了解到BST的中序遍历是递增序列

2.我们用一个临时节点tmp储存p的中序遍历的下一个节点，如果p->right不存在，那么tmp就是从root到p的路径中大于p->val的最小数，否则就遍历p的右子树，找到最左边的节点即可

代码

class Solution {
public:
	/*
	 * @param root: The root of the BST.
	 * @param p: You need find the successor node of p.
	 * @return: Successor of p.
	 */
	TreeNode * tmp;

	TreeNode * inorderSuccessor(TreeNode * root, TreeNode * p) {
		// write your code here'
		if (root == nullptr || p == nullptr) {
			return root;
		}

		tmp = nullptr;
		if (p->right == nullptr) {
			dfs(root, p);
			return tmp;
		}

		p = p->right;
		while (p->left != nullptr) {
			p = p->left;
		}

		return p;
	}

	void dfs(TreeNode * root, TreeNode * p) {
		if (root->val == p->val) {
			return ;
		}

		if (root->val > p->val) {
			tmp = root;
			dfs(root->left, p);
		}

		if (root->val < p->val) {
			dfs(root->right, p);
		}
	}

};