[LeetCode] 404. Sum of Left Leaves

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.

左叶子之和。题意是给一个二叉树，返回所有左孩子的值的和。

这道题有BFS和DFS两种思路。这道题不难，但是无论是BFS还是DFS，记住遍历的时候永远都是看当前节点是否有左孩子，如果有，说明有戏，但是别急着先遍历过去，要先看一下这个左孩子自身是否还有子节点，如果有，其实是不满足题意的。如果当前节点有有孩子，就直接加入queue继续下一轮遍历了，不需要再做额外处理。

BFS

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public int sumOfLeftLeaves(TreeNode root) {
18         // corner case
19         if (root == null) {
20             return 0;
21         }
22         // normal case
23         int res = 0;
24         Queue<TreeNode> queue = new LinkedList<>();
25         queue.offer(root);
26         while (!queue.isEmpty()) {
27             TreeNode cur = queue.poll();
28             if (cur.left != null) {
29                 if (cur.left.left == null && cur.left.right == null) {
30                     res += cur.left.val;
31                 } else {
32                     queue.offer(cur.left);
33                 }
34             }
35             if (cur.right != null) {
36                 queue.offer(cur.right);
37             }
38         }
39         return res;
40     }
41 }

DFS

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public int sumOfLeftLeaves(TreeNode root) {
18         if (root == null) {
19             return 0;
20         }
21         int res = 0;
22         if (root.left != null) {
23             if (root.left.left == null && root.left.right == null) {
24                 res += root.left.val;
25             } else {
26                 res += sumOfLeftLeaves(root.left);
27             }
28         }
29         res += sumOfLeftLeaves(root.right);
30         return res;
31     }
32 }

LeetCode 题目总结