Sum of Left Leaves - LintCode

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

样例

    3
   / \
  9  20
    /  \
   15   7

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

思路
处理结点有左子树,且左子树是叶子结点的情况

#ifndef C1254_H
#define C1254_H
#include<iostream>
using namespace std;
class TreeNode{
public:
    int val;
    TreeNode *left, *right;
    TreeNode(int val){
        this->val = val;
        this->left = this->right = NULL;
    }
};
class Solution {
public:
    /**
    * @param root: t
    * @return: the sum of all left leaves
    */
    int sumOfLeftLeaves(TreeNode * root) {
        // Write your code here
        if (!root)
            return 0;
        //处理结点有左子树,且左子树是叶子结点的情况
        if (root->left&&!root->left->left&&!root->left->right)
        {
            sum += root->left->val;
        }
        sumOfLeftLeaves(root->left);
        sumOfLeftLeaves(root->right);
        return sum;
    }
    int sum = 0;
};
#endif

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转载自blog.csdn.net/zhaohengchuan/article/details/80826186