🌴 二叉树剪枝

📝 题目描述

​ 给你二叉树的根节点root，此外树的每个节点的值要么是0，要么是1。

​ 返回移除了所有不包含1的子树的原二叉树。

​ 节点node的子树为node本身加上所有node的后代。

📋 代码模板

java

/**
 * 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) {

    }
}

typescript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function pruneTree(root: TreeNode | null): TreeNode | null {
    
}

💡 提示

1. 树中节点的数目在范围 $[1,200]$ 内
2. $Node.val$ 为0或1

🚀 示例

🖊️ 题解

java

class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if (root.left == null && root.right == null && root.val == 0) {
            return null;
        }
        return root;
    }
}

typescript

function pruneTree(root: TreeNode | null): TreeNode | null {
    if (root == null) {
        return null;
    }
    root.left = pruneTree(root.left);
    root.right = pruneTree(root.right);
    if (root.left == null && root.right == null && root.val == 0) {
        return null;
    }
    return root;
}

💭 复杂度分析