404. Sum of Left Leaves

最新推荐文章于 2024-12-15 17:30:53 发布

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最新推荐文章于 2024-12-15 17:30:53 发布

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原文链接：http://www.cnblogs.com/bernieloveslife/p/10234249.html

本文介绍了一种算法，用于求解给定二叉树中所有左叶子节点的值之和。通过深度优先搜索（DFS）遍历二叉树，当遇到左叶子节点时累加其值，最终返回总和。示例展示了在一个特定二叉树结构中，左叶子节点分别为9和15，因此返回值为24。

Find the sum of all left leaves in a given binary tree.

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

# 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):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root is None:
            return 0
        res = 0
        def preorder(root):
            nonlocal res
            if root.left:
                if root.left.left is None and root.left.right is None:
                    res += root.left.val
            if root.left:
                preorder(root.left)
            if root.right:
                preorder(root.right)
        preorder(root)
        return res

转载于:https://www.cnblogs.com/bernieloveslife/p/10234249.html