题目描述补充

题目：二叉树的最长连续子序列Ⅰ

给定一棵二叉树，其中每个节点的值都是非负整数。我们需要找到这棵树中所有从根节点到叶子节点的路径中，路径上节点值形成的最长连续递增子序列的长度。请编写一个函数来找出这个最长连续递增子序列的长度。

注意：

• 路径定义为从树的根节点开始，到任意一个叶子节点结束的路径。
• 叶子节点是没有子节点的节点。
• 如果不存在这样的路径，返回0。

示例

考虑以下二叉树：

    1
   / \
  2   3
 / \   \
2   3   4
   \
    5

最长连续递增子序列是 [1, 2, 3, 4]，所以返回4。

PHP 示例代码

<?php

class TreeNode {
    public $val;
    public $left;
    public $right;
    
    function __construct($val = 0, $left = null, $right = null) {
        $this->val = $val;
        $this->left = $left;
        $this->right = $right;
    }
}

function findLongestContinuousIncreasingPath($root) {
    if ($root === null) return 0;
    return dfs($root, $root->val, 1);
}

function dfs($node, $prevVal, $currLength) {
    if ($node === null) return 0;
    
    $maxLength = $currLength;
    
    if ($node->val > $prevVal) {
        $maxLength = max($maxLength, dfs($node->left, $node->val, $currLength + 1));
        $maxLength = max($maxLength, dfs($node->right, $node->val, $currLength + 1));
    } else {
        $maxLength = max($maxLength, dfs($node->left, $node->val, 1));
        $maxLength = max($maxLength, dfs($node->right, $node->val, 1));
    }
    
    return $maxLength;
}

// 使用示例
$root = new TreeNode(1);
$root->left = new TreeNode(2);
$root->right = new TreeNode(3);
$root->left->left = new TreeNode(2);
$root->left->right = new TreeNode(3);
$root->right->right = new TreeNode(4);
$root->left->right->right = new TreeNode(5);

echo findLongestContinuousIncreasingPath($root);  // 输出 4
?>

Python 示例代码

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def findLongestContinuousIncreasingPath(root):
    if not root:
        return 0
    return dfs(root, root.val, 1)

def dfs(node, prev_val, curr_length):
    if not node:
        return 0
    
    max_length = curr_length
    
    if node.val > prev_val:
        max_length = max(max_length, dfs(node.left, node.val, curr_length + 1))
        max_length = max(max_length, dfs(node.right, node.val, curr_length + 1))
    else:
        max_length = max(max_length, dfs(node.left, node.val, 1))
        max_length = max(max_length, dfs(node.right, node.val, 1))
    
    return max_length

# 使用示例
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(2)
root.left.right = TreeNode(3)
root.right.right = TreeNode(4)
root.left.right.right = TreeNode(5)

print(findLongestContinuousIncreasingPath(root))  # 输出 4

JavaScript 示例代码

function TreeNode(val, left, right) {
    this.val = (val===undefined ? 0 : val)
    this.left = (left===undefined ? null : left)
    this.right = (right===undefined ? null : right)
}

function findLongestContinuousIncreasingPath(root) {
    if (!root) return 0;
    return dfs(root, root.val, 1);
}

function dfs(node, prevVal, currLength) {
    if (!node) return 0;
    
    let maxLength = currLength;
    
    if (node.val > prevVal) {
        maxLength = Math.max(maxLength, dfs(node.left, node.val, currLength + 1));
        maxLength = Math.max(maxLength, dfs(node.right, node.val, currLength + 1));
    } else {
        maxLength = Math.max(maxLength, dfs(node.left, node.val, 1));
        maxLength = Math.max(maxLength, dfs(node.right, node.val, 1));
    }
    
    return maxLength;
}

// 使用示例
let root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(2);
root.left.right = new TreeNode(3);
root.right.right = new TreeNode(4);
root.left.right.right = new TreeNode(5);

console.log(findLongestContinuousIncreasingPath(root));  // 输出 4

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