103 lines
2.4 KiB
Markdown
103 lines
2.4 KiB
Markdown
---
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layout: page
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title: Merge Two Binary Trees
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last_solved:
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category: trees
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leetcode_url: https://leetcode.com/problems/merge-two-binary-trees
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status: Attempted
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---
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Problem
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-------
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```
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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.
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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.
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Example 1:
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Input:
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Tree 1 Tree 2
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1 2
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/ \ / \
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3 2 1 3
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/ \ \
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5 4 7
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Output:
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Merged tree:
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3
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/ \
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4 5
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/ \ \
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5 4 7
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```
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Solution
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----------
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{% highlight python %}
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def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
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def dfs(t1, t2, t3):
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# sums up the roots
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t3.val = t1.val + t2.val
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# if only t1 has left child
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if t1.left and not t2.left:
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t3.left = t1.left
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# if only t1 has right child
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if t1.right and not t2.right:
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t3.right = t1.right
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# if only t2 has left child
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if t2.left and not t1.left:
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t3.left = t2.left
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# if only t2 has right child
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if t2.right and not t1.right:
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t3.right = t2.right
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# recurse down left if both have left
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if t1.left and t2.left:
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t3.left = TreeNode(0)
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dfs(t1.left, t2.left, t3.left)
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if t1.right and t2.right:
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t3.right = TreeNode(0)
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dfs(t1.right, t2.right, t3.right)
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# start of main
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if not t1 and not t2:
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return
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if t1 and not t2:
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return t1
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if t2 and not t1:
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return t2
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# if t1 and t2
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t3 = TreeNode(0)
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dfs(t1, t2, t3)
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return t3
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{% endhighlight %}
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______________
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{% highlight python %}
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{% endhighlight %}
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![image1]() |