LintCode 1126: Merge Two Binary Trees (二叉树经典题)

原创已于 2024-01-02 02:41:55 修改 · 243 阅读

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于 2019-01-01 13:19:39 首次发布

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本文介绍了一种将两棵二叉树合并成一棵新二叉树的算法，当两个节点重叠时，新节点的值为其和。通过递归方式处理根节点及左右子树，实现了树的完全合并。

Merge Two Binary Trees
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
Merged tree:
3
/
4 5
/ \ \
5 4 7
Notice
The merging process must start from the root nodes of both trees.

思路：
二叉树递归。
注意：

Tree * node = NULL; //记得这里要初始化为NULL, 否则当t1, t2都是NULL时，返回unknown。
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 t1: the root of the first tree
     * @param t2: the root of the second tree
     * @return: the new binary tree after merge
     */
    TreeNode * mergeTrees(TreeNode * t1, TreeNode * t2) {
        TreeNode * node = NULL;
        if (t1 && t2) {
            node = new TreeNode(t1->val + t2->val);
            node->left = mergeTrees(t1->left, t2->left);
            node->right = mergeTrees(t1->right, t2->right);
        } else if (t1 && !t2) {
            node = new TreeNode(t1->val);
            node->left = mergeTrees(t1->left, NULL);
            node->right = mergeTrees(t1->right, NULL);
        } else if (!t1 && t2) {
            node = new TreeNode(t2->val);
            node->left = mergeTrees(NULL, t2->left);
            node->right = mergeTrees(NULL, t2->right);
        }     
        return node;
    }
};

二刷：

/**
 * 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 t1: the root of the first tree
     * @param t2: the root of the second tree
     * @return: the new binary tree after merge
     */
    TreeNode* mergeTrees(TreeNode *t1, TreeNode *t2) {
        if (!t1 && !t2) return NULL;
        TreeNode *newNode = NULL;
        if (!t1) {
            newNode = new TreeNode(t2->val);
        } else if (!t2) {
            newNode = new TreeNode(t1->val);
        } else {
            newNode = new TreeNode(t1->val + t2->val);
        }
        newNode->left = mergeTrees(t1 ? t1->left : NULL, t2 ? t2->left : NULL);
        newNode->right = mergeTrees(t1 ? t1->right : NULL, t2 ? t2->right : NULL);
        return newNode;
    }
};