404. Sum of Left Leaves*

404. Sum of Left Leaves*

https://leetcode.com/problems/sum-of-left-leaves/

题目描述

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.

C++ 实现 1

注意是求所有 left 叶子节点的值的和. 如果使用层序遍历.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root) return 0;
        int sum = 0;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            auto r = q.front();
            q.pop();
            if (r->left) {
                if (!r->left->left && !r->left->right) sum += r->left->val;
                q.push(r->left);
            } 
            if (r->right) q.push(r->right);
        }
        return sum;
    }
};

C++ 实现 2

使用递归来做.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root || (!root->left && !root->right))
            return 0;

        if (root->left && (!root->left->left && !root->left->right))
            return root->left->val + sumOfLeftLeaves(root->right);
        else
            return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
};