将给定两个二叉树,想象当你将它们中的一个覆盖到另一个上时,两个二叉树的一些节点便会重叠。
你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠,那么将他们的值相加作为节点合并后的新值,否则不为 NULL 的节点将直接作为新二叉树的节点。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-two-binary-trees
这里可以使用迭代和递归两种方式来解决。时间和空间复杂度都是O(N)。
首先我们使用递归。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if t1 is None:
return t2
if t2 is None:
return t1
t1.val += t2.val
t1.left = self.mergeTrees(t1.left, t2.left)
t1.right = self.mergeTrees(t1.right, t2.right)
return t1
接下来是迭代。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null)
return t2;
Stack < TreeNode[] > stack = new Stack < > ();
stack.push(new TreeNode[] {t1, t2});
while (!stack.isEmpty()) {
TreeNode[] t = stack.pop();
if (t[0] == null || t[1] == null) {
continue;
}
t[0].val += t[1].val;
if (t[0].left == null) {
t[0].left = t[1].left;
} else {
stack.push(new TreeNode[] {t[0].left, t[1].left});
}
if (t[0].right == null) {
t[0].right = t[1].right;
} else {
stack.push(new TreeNode[] {t[0].right, t[1].right});
}
}
return t1;
}
}
作者:LeetCode
链接:https://leetcode-cn.com/problems/merge-two-binary-trees/solution/he-bing-er-cha-shu-by-leetcode/
来源:力扣(LeetCode)