LintCode 1003: Binary Tree Pruning (二叉树经典题)

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博客围绕二叉树剪枝问题展开，给定节点值为0或1的二叉树，需移除不包含1的子树。给出多个输入输出示例，还介绍了递归解法，强调添加特定代码行可避免考虑指针的指针或引用。

1003. Binary Tree Pruning

中文English

Given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1.

Return this 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.)

Example

Example 1:

Input: {1,#,0,0,1}

Output: {1,#,0,#,1}

Explanation:

Only the red nodes satisfy the property "every subtree not containing a 1".

The diagram on the right represents the answer.

Example 2:

Input: {1,0,1,0,0,0,1}

Output: {1,#,1,#,1}

Example 3:

Input: {1,1,0,1,1,0,1,0}

Output: {1,1,0,1,1,#,1}

Notice

The binary tree will have at most 100 nodes.

The value of each node will only be 0 or 1.

解法1：递归。

这题要注意加这2行：
root->left = leftNode;
root->right =rightNode;
这样就不用考虑指针的指针或指针的引用了。

代码如下:

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: the root
     * @return: the same tree where every subtree (of the given tree) not containing a 1 has been removed
     */
    TreeNode * pruneTree(TreeNode * root) {
        if (!root) return NULL;
        
        TreeNode * leftNode = pruneTree(root->left);
        TreeNode * rightNode = pruneTree(root->right);
        
        root->left = leftNode;
        root->right = rightNode;
        
        if (!leftNode && !rightNode && root->val == 0) {
            root = NULL;
        }
        return root;
    }
};