验证二叉树的前序序列化

rust 解法, 执行用时: 0 ms, 内存消耗: 2.2 MB, 提交时间: 2024-03-31 00:07:53

impl Solution {
    pub fn is_valid_serialization(preorder: String) -> bool {
        let mut nodes = 0;
        let preorder: Vec<&str> = preorder.split_terminator(",").collect();
        for i in 0..preorder.len() {
            if preorder[i] != "#" {
                nodes += 1;
            } else {
                nodes -= 1;
            }
            if i < preorder.len() - 1 && nodes < 0 { return false; }
        }
        nodes == -1
    }
}

rust 解法, 执行用时: 0 ms, 内存消耗: 2.1 MB, 提交时间: 2024-03-31 00:07:33

impl Solution {
    pub fn is_valid_serialization(preorder: String) -> bool {
        let mut stack = Vec::new();
        let preorder: Vec<&str> = preorder.split_terminator(",").collect();
        for s in preorder.into_iter() {
            if s != "#" {
                stack.push(s);
                continue;
            }
            while !stack.is_empty() && stack[stack.len() - 1] == "#" {
                stack.pop();
                if let Some(x) = stack.pop() {
                    if x == "#" { return false; }
                } else {
                    return false;
                }
            }
            stack.push("#");
        }
        stack.len() == 1 && stack[0] == "#"
    }
}

golang 解法, 执行用时: 4 ms, 内存消耗: 2 MB, 提交时间: 2023-06-13 09:49:37

func isValidSerialization(preorder string) bool {
    n := len(preorder)
    slots := 1
    for i := 0; i < n; {
        if slots == 0 {
            return false
        }
        if preorder[i] == ',' {
            i++
        } else if preorder[i] == '#' {
            slots--
            i++
        } else {
            // 读一个数字
            for i < n && preorder[i] != ',' {
                i++
            }
            slots++ // slots = slots - 1 + 2
        }
    }
    return slots == 0
}

javascript 解法, 执行用时: 68 ms, 内存消耗: 41.3 MB, 提交时间: 2023-06-13 09:49:23

/**
 * @param {string} preorder
 * @return {boolean}
 */
var isValidSerialization = function(preorder) {
    const n = preorder.length;
    let i = 0;
    let slots = 1;
    while (i < n) {
        if (slots === 0) {
            return false;
        }
        if (preorder[i] === ',') {
            ++i;
        } else if (preorder[i] === '#') {
            --slots;
            ++i;
        } else {
            // 读一个数字
            while (i < n && preorder[i] !== ',') {
                ++i;
            }
            ++slots; // slots = slots - 1 + 2
        }
    }
    return slots === 0;
};

java 解法, 执行用时: 2 ms, 内存消耗: 39.9 MB, 提交时间: 2023-06-13 09:49:11

class Solution {
    public boolean isValidSerialization(String preorder) {
        int n = preorder.length();
        int i = 0;
        int slots = 1;
        while (i < n) {
            if (slots == 0) {
                return false;
            }
            if (preorder.charAt(i) == ',') {
                i++;
            } else if (preorder.charAt(i) == '#'){
                slots--;
                i++;
            } else {
                // 读一个数字
                while (i < n && preorder.charAt(i) != ',') {
                    i++;
                }
                slots++; // slots = slots - 1 + 2
            }
        }
        return slots == 0;
    }
}

cpp 解法, 执行用时: 0 ms, 内存消耗: 6.5 MB, 提交时间: 2023-06-13 09:48:58

class Solution {
public:
    bool isValidSerialization(string preorder) {
        int n = preorder.length();
        int i = 0;
        int slots = 1;
        while (i < n) {
            if (slots == 0) {
                return false;
            }
            if (preorder[i] == ',') {
                i++;
            } else if (preorder[i] == '#'){
                slots--;
                i++;
            } else {
                // 读一个数字
                while (i < n && preorder[i] != ',') {
                    i++;
                }
                slots++; // slots = slots - 1 + 2
            }
        }
        return slots == 0;
    }
};

python3 解法, 执行用时: 40 ms, 内存消耗: 16.1 MB, 提交时间: 2023-06-13 09:48:36

# 计算入度出度
class Solution:
    def isValidSerialization(self, preorder: str) -> bool:
        nodes = preorder.split(',')
        diff = 1
        for node in nodes:
            diff -= 1
            if diff < 0:
                return False
            if node != '#':
                diff += 2
        return diff == 0

python3 解法, 执行用时: 52 ms, 内存消耗: 16 MB, 提交时间: 2023-06-13 09:47:41

class Solution:
    def isValidSerialization(self, preorder: str) -> bool:
        stack = list()
        for i in preorder.split(','):
            stack.append(i)
            while len(stack) > 2 and stack[-1] == '#' and stack[-2] == '#' and stack[-3].isdigit():
                stack.pop()
                stack.pop()
                stack.pop()
                stack.append('#')
        return True if stack == ['#'] else False

golang 解法, 执行用时: 0 ms, 内存消耗: 2 MB, 提交时间: 2023-06-13 09:43:54

func isValidSerialization(preorder string) bool {
    n := len(preorder)
    stk := []int{1}
    for i := 0; i < n; {
        if len(stk) == 0 {
            return false
        }
        if preorder[i] == ',' {
            i++
        } else if preorder[i] == '#' {
            stk[len(stk)-1]--
            if stk[len(stk)-1] == 0 {
                stk = stk[:len(stk)-1]
            }
            i++
        } else {
            // 读一个数字
            for i < n && preorder[i] != ',' {
                i++
            }
            stk[len(stk)-1]--
            if stk[len(stk)-1] == 0 {
                stk = stk[:len(stk)-1]
            }
            stk = append(stk, 2)
        }
    }
    return len(stk) == 0
}

javascript 解法, 执行用时: 60 ms, 内存消耗: 41.5 MB, 提交时间: 2023-06-13 09:43:38

/**
 * @param {string} preorder
 * @return {boolean}
 */
var isValidSerialization = function(preorder) {
    const n = preorder.length;
    let i = 0;
    const stack = [1];
    while (i < n) {
        if (!stack.length) {
            return false;
        }
        if (preorder[i] === ',') {
            ++i;
        } else if (preorder[i] === '#') {
            stack[stack.length - 1]--;
            if (stack[stack.length - 1] === 0) {
                stack.pop();
            } 
            ++i;
        } else {
            // 读一个数字
            while (i < n && preorder[i] !== ',') {
                ++i;
            }
            stack[stack.length - 1]--;
            if (stack[stack.length - 1] === 0) {
                stack.pop();
            }
            stack.push(2);
        }
    }
    return stack.length === 0;
};

java 解法, 执行用时: 4 ms, 内存消耗: 40.5 MB, 提交时间: 2023-06-13 09:43:23

class Solution {
    public boolean isValidSerialization(String preorder) {
        int n = preorder.length();
        int i = 0;
        Deque<Integer> stack = new LinkedList<Integer>();
        stack.push(1);
        while (i < n) {
            if (stack.isEmpty()) {
                return false;
            }
            if (preorder.charAt(i) == ',') {
                i++;
            } else if (preorder.charAt(i) == '#'){
                int top = stack.pop() - 1;
                if (top > 0) {
                    stack.push(top);
                }
                i++;
            } else {
                // 读一个数字
                while (i < n && preorder.charAt(i) != ',') {
                    i++;
                }
                int top = stack.pop() - 1;
                if (top > 0) {
                    stack.push(top);
                }
                stack.push(2);
            }
        }
        return stack.isEmpty();
    }
}

cpp 解法, 执行用时: 8 ms, 内存消耗: 6.6 MB, 提交时间: 2023-06-13 09:43:07

// 基于栈
class Solution {
public:
    bool isValidSerialization(string preorder) {
        int n = preorder.length();
        int i = 0;
        stack<int> stk;
        stk.push(1);
        while (i < n) {
            if (stk.empty()) {
                return false;
            }
            if (preorder[i] == ',') {
                i++;
            } else if (preorder[i] == '#'){
                stk.top() -= 1;
                if (stk.top() == 0) {
                    stk.pop();
                }
                i++;
            } else {
                // 读一个数字
                while (i < n && preorder[i] != ',') {
                    i++;
                }
                stk.top() -= 1;
                if (stk.top() == 0) {
                    stk.pop();
                }
                stk.push(2);
            }
        }
        return stk.empty();
    }
};

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