algorithm notes: leetcode#617 merge two binary trees

Problem

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 1:

Input:
            Tree 1                    Tree 2
                1                          2
              /                         /    
            3      2                 1      3
           /                                    
         5                               4      7
Output:
Merged tree:
                        3
                      /    
                    4      5
                  /           
                5      4      7

Note: The merging process must start from the root nodes of both trees.

Solution

Basic idea

We can solve this problem with recursion. If one of the two trees is None, just return the tree which is not None. If both are not None, add their values as value of merged node. Then merge t1.left and t2.left with function mergeTrees itself as the left child node of merged node. The same for t1.right and t2.right.

Python implementation

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#class TreeNode(object):
#    def __init__(self, x):
#        self.val = x
#        self.left = None
#        self.right = None
class :
    def mergeTrees(self, t1, t2):
        """
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        """
        if t1 is None:
            return t2
        if t2 is not None:
            t1.val += t2.val
            t1.left = self.mergeTrees(t1.left, t2.left)
            t1.right = self.mergeTrees(t1.right, t2.right)
        return t1

Java implementation

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null){ return t2; }
        if(t2 != null){
            t1.val += t2.val;
            t1.left = mergeTrees(t1.left, t2.left);
            t1.right = mergeTrees(t1.right, t2.right);
        }
        return t1;
    }
}

Time complexity analysis

O(N). N is the minimum number of nodes from two trees. We traverse all nodes of the tree with fewer nodes.

Space complexity analysis

O(N).

Link

617. Merge Two Binary Trees
617. 合并二叉树
(中文版) 算法笔记: 力扣#617 合并二叉树