404. 左叶子之和

English Version

题目描述

给定二叉树的根节点 root ，返回所有左叶子之和。

 

示例 1：

输入: root = [3,9,20,null,null,15,7] 
输出: 24 
解释: 在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24

示例 2:

输入: root = [1]
输出: 0

 

提示:

• 节点数在 [1, 1000] 范围内
• -1000 <= Node.val <= 1000

 

解法

Python3

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


class Solution:
    def sumOfLeftLeaves(self, root: TreeNode) -> int:
        if root is None:
            return 0
        res = 0
        if root.left and root.left.left is None and root.left.right is None:
            res += root.left.val
        res += self.sumOfLeftLeaves(root.left)
        res += self.sumOfLeftLeaves(root.right)
        return res

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int res = 0;
        if (root.left != null && root.left.left == null && root.left.right == null) {
            res += root.left.val;
        }
        res += sumOfLeftLeaves(root.left);
        res += sumOfLeftLeaves(root.right);
        return res;
    }
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

const dfs = (root: TreeNode | null, isLeft: boolean) => {
    let res = 0;
    const { val, left, right } = root;
    if (left == null && right == null) {
        if (isLeft) {
            return val;
        }
        return res;
    }
    if (left != null) {
        res += dfs(left, true);
    }
    if (right != null) {
        res += dfs(right, false);
    }
    return res;
};

function sumOfLeftLeaves(root: TreeNode | null): number {
    return dfs(root, false);
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, is_left: bool) -> i32 {
        let node = root.as_ref().unwrap().borrow();
        let left = &node.left;
        let right = &node.right;
        let mut res = 0;
        if left.is_none() && right.is_none() {
            if is_left {
                return node.val;
            }
            return res;
        }
        if left.is_some() {
            res += Self::dfs(left, true);
        }
        if right.is_some() {
            res += Self::dfs(right, false);
        }
        res
    }

    pub fn sum_of_left_leaves(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        Self::dfs(&root, false)
    }
}

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