[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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        // corner case
        if (root == null) {
            return 0;
        }
        // normal case
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur.left != null) {
                if (cur.left.left == null && cur.left.right == null) {
                    res += cur.left.val;
                } else {
                    queue.offer(cur.left);
                }
            }
            if (cur.right != null) {
                queue.offer(cur.right);
            }
        }
        return res;
    }
}

DFS

时间O(n)

空间O(n)

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int res = 0;
        if (root.left != null) {
            if (root.left.left == null && root.left.right == null) {
                res += root.left.val;
            } else {
                res += sumOfLeftLeaves(root.left);
            }
        }
        res += sumOfLeftLeaves(root.right);
        return res;
    }
}

LeetCode 题目总结