Description

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.

分析

题目的意思是:求二叉树左叶子节点的和。

  • 树模型一般用递归的方法,终止条件当然是该节点为空。如果左子节点存在,且左叶子节点没有子节点,那么该左节点为题目要求的节点。然后继续递归找出所有的左叶子节点。

代码

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

return sumOfLeftLeaves(root->left)+sumOfLeftLeaves(root->right);
}
};

参考文献

​[LeetCode] Sum of Left Leaves 左子叶之和​