提示：文章写完后，目录可以自动生成，如何生成可参考右边的帮助文档

---

前言

---

这道题的难点在于要一直剪枝，直到没有值为 0 的叶子节点为止，只有从后序遍历位置自底向上处理才能获得最高的效率

一、力扣814. 二叉树剪枝

/*** 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) {if(root == null){return null;}root.left = pruneTree(root.left);root.right = pruneTree(root.right);if(root.left == null && root.right == null){if(root.val == 0){return null;}}return root;}
}

二、力扣1325. 删除给定值的叶子节点

/*** 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 removeLeafNodes(TreeNode root, int target) {if(root == null){return null;}TreeNode left = removeLeafNodes(root.left,target);TreeNode right = removeLeafNodes(root.right,target);if(left == null && right == null){if(root.val == target){return null;}}root.left = left;root.right = right;return root;}
}