力扣labuladong——一刷day70

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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;
    }
}