9/19刷题左叶子之和

1.左叶子之和


思路一：DFS

/**
 * 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:
    bool isLeafNode(TreeNode* node) {
        return !node->left && !node->right;
    }//判断是不是叶子结点

    int dfs(TreeNode* node) {
        int ans = 0;
        if (node->left) {
            ans += isLeafNode(node->left) ? node->left->val : dfs(node->left);//找到叶子结点就加一个值，否则继续往左找
        }//左面递归寻找
        if (node->right && !isLeafNode(node->right)) {
            ans += dfs(node->right);
        }//右面递归寻找
        return ans;
    }

    int sumOfLeftLeaves(TreeNode* root) {
        return root ? dfs(root) : 0;
    }
};