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算法笔记: 力扣#404 左叶子之和

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def dfs(node, is_left):
            if not node:
                return 0
            if not node.left and not node.right:
                if is_left:
                    return node.val
                return 0
            return dfs(node.left, True) + dfs(node.right, False)
        
        return dfs(root, False)

Java 实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    private int dfs(TreeNode node, boolean isLeft){
        if(node == null){ return 0; }
        if(node.left == null && node.right == null){
            if(isLeft){ return node.val; }
            return 0;
        }
        return dfs(node.left, true) + dfs(node.right, false);
    }
    
    public int sumOfLeftLeaves(TreeNode root) {
        return dfs(root, false);
    }
}

时间复杂度

O(n).

空间复杂度

O(1).

链接

404. Sum of Left Leaves
404. 左叶子之和
 (English version) Algorithm Notes: Leetcode#404 Sum of Left Leaves