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面试题 04.06. 后继者

English Version

题目描述

设计一个算法，找出二叉搜索树中指定节点的“下一个”节点（也即中序后继）。

如果指定节点没有对应的“下一个”节点，则返回null。

示例 1:

输入: root = [2,1,3], p = 1

  2
 / \
1   3

输出: 2

示例 2:

输入: root = [5,3,6,2,4,null,null,1], p = 6

      5
     / \
    3   6
   / \
  2   4
 /   
1

输出: null

解法

方法一

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> TreeNode:
        def dfs(root):
            if root is None:
                return
            dfs(root.left)
            nonlocal ans, prev
            if prev == p:
                ans = root
            prev = root
            dfs(root.right)

        ans = prev = None
        dfs(root)
        return ans

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode prev;
    private TreeNode p;
    private TreeNode ans;

    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        prev = null;
        ans = null;
        this.p = p;
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        if (prev == p) {
            ans = root;
        }
        prev = root;
        dfs(root.right);
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* prev;
    TreeNode* p;
    TreeNode* ans;

    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        this->p = p;
        dfs(root);
        return ans;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->left);
        if (prev == p) ans = root;
        prev = root;
        dfs(root->right);
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderSuccessor(root *TreeNode, p *TreeNode) *TreeNode {
	var prev, ans *TreeNode
	var dfs func(root *TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		if prev == p {
			ans = root
		}
		prev = root
		dfs(root.Right)
	}
	dfs(root)
	return ans
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @return {TreeNode}
 */
var inorderSuccessor = function (root, p) {
    if (root == null) {
        return root;
    }
    const { val, left, right } = root;
    const res = inorderSuccessor(left, p);
    if (res != null) {
        return res;
    }
    if (val > p.val) {
        return root;
    }
    return inorderSuccessor(right, p);
};

方法二

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        stack<TreeNode*> stk;
        TreeNode* cur = root;
        while (cur != nullptr || !stk.empty()) {
            if (cur == nullptr) {
                cur = stk.top();
                stk.pop();
                if (cur->val > p->val) {
                    return cur;
                }
                cur = cur->right;
            } else {
                stk.push(cur);
                cur = cur->left;
            }
        }
        return cur;
    }
};

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @return {TreeNode}
 */
var inorderSuccessor = function (root, p) {
    const stack = [];
    let cur = root;
    while (cur != null || stack.length !== 0) {
        if (cur == null) {
            cur = stack.pop();
            if (cur.val > p.val) {
                return cur;
            }
            cur = cur.right;
        } else {
            stack.push(cur);
            cur = cur.left;
        }
    }
    return cur;
};