[LeetCode] 298. Binary Tree Longest Consecutive Sequence 二叉樹最長連續序列

Given a binary tree, find the length of the longest consecutive sequence path.

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).

For example,

1
    \
     3
    / \
   2   4
        \
         5

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

2
    \
     3
    / 
   2    
  / 
 1

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

求二叉樹的最長連續序列的長度，要從父節點到子節點。最長連續子序列必須是從root到leaf的方向。 好比 1->2，返回長度2， 好比1->3->4->5，返回3->4->5這個子序列的長度3。

解法：遞歸遍歷binary tree，遞歸函數傳入父節點的值，以幫助子節點判斷是否連續。

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

/**
 * 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:

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++:

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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