[LeetCode] 298. Binary Tree Longest Consecutive Sequence 二叉树最长连续序列

2021年09月15日 阅读数:2
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Given a binary tree, find the length of the longest consecutive sequence path.html

The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from parent to child (cannot be the reverse).java

For example,node

   1
    \
     3
    / \
   2   4
        \
         5

Longest consecutive sequence path is 3-4-5, so return 3.python

   2
    \
     3
    / 
   2    
  / 
 1

Longest consecutive sequence path is 2-3,not3-2-1, so return 2.函数

求二叉树的最长连续序列的长度,要从父节点到子节点。最长连续子序列必须是从root到leaf的方向。 好比 1->2,返回长度2, 好比1->3->4->5,返回3->4->5这个子序列的长度3。post

解法:递归遍历binary tree,递归函数传入父节点的值,以帮助子节点判断是否连续。url

Java: Time Complexity - O(n),  Space Complexity - O(n)spa

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int maxLen = 0;
    
    public int longestConsecutive(TreeNode root) {
        longestConsecutive(root, 0, 0);
        return maxLen;
    }
    
    private void longestConsecutive(TreeNode root, int lastVal, int curLen) {
        if (root == null) return;
        if (root.val != lastVal + 1) curLen = 1;
        else curLen++;
        maxLen = Math.max(maxLen, curLen);
        longestConsecutive(root.left, root.val, curLen);
        longestConsecutive(root.right, root.val, curLen);
    }
}

Python:code

class Solution(object):
    def longestConsecutive(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.max_len = 0

        def longestConsecutiveHelper(root):
            if not root:
                return 0
    
            left_len = longestConsecutiveHelper(root.left)
            right_len = longestConsecutiveHelper(root.right)
    
            cur_len = 1
            if root.left and root.left.val == root.val + 1:
                cur_len = max(cur_len, left_len + 1);
            if root.right and root.right.val == root.val + 1:
                cur_len = max(cur_len, right_len + 1)

            self.max_len = max(self.max_len, cur_len, left_len, right_len)

            return cur_len

        longestConsecutiveHelper(root)
        return self.max_len

C++:htm

class Solution {
public:
    int longestConsecutive(TreeNode* root) {
        if (!root) return 0;
        int res = 0;
        dfs(root, root->val, 0, res);
        return res;
    }
    void dfs(TreeNode *root, int v, int out, int &res) {
        if (!root) return;
        if (root->val == v + 1) ++out;
        else out = 1;
        res = max(res, out);
        dfs(root->left, root->val, out, res);
        dfs(root->right, root->val, out, res);
    }
};

C++:

class Solution {
public:
    int longestConsecutive(TreeNode* root) {
        if (!root) return 0;
        int res = 0;
        dfs(root, 1, res);
        return res;
    }
    void dfs(TreeNode *root, int len, int &res) {
        res = max(res, len);
        if (root->left) {
            if (root->left->val == root->val + 1) dfs(root->left, len + 1, res);
            else dfs(root->left, 1, res);
        }
        if (root->right) {
            if (root->right->val == root->val + 1) dfs(root->right, len + 1, res);
            else dfs(root->right, 1, res);
        }
    }
};

C++:  

class Solution {
public:
    int longestConsecutive(TreeNode* root) {
        return helper(root, NULL, 0);
    }
    int helper(TreeNode *root, TreeNode *p, int res) {
        if (!root) return res;
        res = (p && root->val == p->val + 1) ? res + 1 : 1;
        return max(res, max(helper(root->left, root, res), helper(root->right, root, res)));
    }
};

C++: 迭代  

class Solution {
public:
    int longestConsecutive(TreeNode* root) {
        if (!root) return 0;
        int res = 0;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            int len = 1;
            TreeNode *t = q.front(); q.pop();
            while ((t->left && t->left->val == t->val + 1) || (t->right && t->right->val == t->val + 1)) {
                if (t->left && t->left->val == t->val + 1) {
                    if (t->right) q.push(t->right);
                    t = t->left;
                } else if (t->right && t->right->val == t->val + 1) {
                    if (t->left) q.push(t->left);
                    t = t->right;
                }
                ++len;
            }
            if (t->left) q.push(t->left);
            if (t->right) q.push(t->right);
            res = max(res, len);
        }
        return res;
    }
};

  

  

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