404. Sum of Left Leaves(统计左叶子节点的和)

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.

方法一：递归调用
时间复杂度：o（n）                   空间复杂度：o（1）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if(root==null) return 0;
        if(isLeft(root.left)) return root.left.val+sumOfLeftLeaves(root.right); //找到左子叶之后继续找
        return sumOfLeftLeaves(root.left)+sumOfLeftLeaves(root.right); 
    }
    private boolean isLeft(TreeNode root){
        if(root==null) return false;
        return root.left==null&&root.right==null;  //子叶是没有左右孩子的结点
    }
}