LeetCode 814. Binary Tree Pruning

Problem

LeetCode 814. Binary Tree Pruning

1. 题目简述

给出一颗二叉树，其每个节点都由0或1组成，删除其所有的节点都为0的子树，返回新的树的root节点。例如：

Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]

2. 算法思路

递归：

和以前的一些求子树和的题目很像，例如LeetCode 508. Most Frequent Subtree Sum，我们也可以求子树和，如果和为0，则说明将当前节点置为null即可。与返回布尔值的效果一致。仍然使用后序遍历。

3. 解法

1. 如果节点为null，直接返回0.

/**
 * 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 TreeNode pruneTree(TreeNode root) {
        int rootSum = subtreeSum(root);
        if (rootSum == 0) {
            return null;
        }

        return root;
    }

    private int subtreeSum(TreeNode root){
        if (root == null) {
            return 0;
        }

        int leftSum = subtreeSum(root.left);
        int rightSum = subtreeSum(root.right);

        if (leftSum == 0) {
            root.left = null;
        }
        if (rightSum == 0) {
            root.right = null;
        }

        return root.val + leftSum + rightSum;
    }
}