814-Binary Tree Pruning

Description: 
We are given the head node root of a binary tree, where additionally every node’s value is either a 0 or a 1.

Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)

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Note:

• The binary tree will have at most 100 nodes
• The value of each node will only be 0 or 1.

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问题描述: 
给定二叉树的头节点root，这个二叉树的每个节点值为0或者1. 
现在要对该二叉树剪枝，对节点剪枝的条件为该节点的所有子树都为0

问题分析: 
利用后序遍历将不符合条件的节点删除就可以了。

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)    root = null;

        return root;
    }
}