LeetCode 617. Merge Two Binary Trees

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题目

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.

答案:

golang版是自己写的,其他的是引用cain_huang 的,文末附上链接

golang版,附带测试

mergeTrees.go

package _617_mergeTrees

import "fmt"

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type TreeNode struct {
     Val int
     Left *TreeNode
     Right *TreeNode
 }

func InitNode(val int, left *TreeNode, right *TreeNode) (ret *TreeNode){
    ret = new(TreeNode)
    ret.Val = val
    ret.Left = left
    ret.Right = right

    return ret
}

func TreePrint(t1 *TreeNode) {

    if nil == t1 {
        fmt.Printf("null, ")
        return
    } else {
        fmt.Printf("%+v, ", t1.Val)
    }

    TreePrint(t1.Left)
    TreePrint(t1.Right)
}


func MergeTrees(t1 *TreeNode, t2 *TreeNode) (t3 *TreeNode) {

    if nil == t1 && nil == t2 {
        return nil
    }
    
    if nil != t1 && nil == t2{
        return t1
    }else if nil == t1 && nil != t2 {
        return t2
    } else {
        t1.Val += t2.Val
        t1.Left = MergeTrees(t1.Left, t2.Left)
        t1.Right = MergeTrees(t1.Right, t2.Right)
    }

    return t1
}

mergeTrees_test.go

package _617_mergeTrees

import (
    "testing"
    "fmt"
)


func treeEqual(t1 *TreeNode, t2 *TreeNode) bool{
    if t1.Val != t2.Val {
        return false
    }

    if nil == t1.Left && nil != t2.Left {
        return false
    }else if nil != t1.Left && nil == t2.Left {
        return false
    } else if nil != t1.Left && nil != t2.Left {
        left := treeEqual(t1.Left, t2.Left)
        if !left {
            return false
        }
    }

    if nil == t1.Right && nil != t2.Right {
        return false
    }else if nil != t1.Right && nil == t2.Right {
        return false
    }else if nil != t1.Right && nil != t2.Right {
        right := treeEqual(t1.Right, t2.Right)
        if !right {
            return false
        }
    }

    return true
}

func Test_mergeTrees(t *testing.T) {
    t1l2 := InitNode(5, nil, nil)
    t1l1 := InitNode(3, t1l2, nil)
    t1r1 := InitNode(2, nil, nil)
    t1 := InitNode(1, t1l1, t1r1)

    t2l2r1 := InitNode(4, nil, nil)
    t2l2r2 := InitNode(7, nil, nil)
    t2l1r1 := InitNode(1, nil, t2l2r1)
    t2l1r2 := InitNode(3, nil, t2l2r2)
    t2 := InitNode(2, t2l1r1, t2l1r2)

    t3l2r1 := InitNode(5, nil, nil)
    t3l2r2 := InitNode(4, nil, nil)
    t3l2r3 := InitNode(7, nil, nil)

    t3l1r1 := InitNode(4, t3l2r1, t3l2r2)
    t3l1r2 := InitNode(5, nil, t3l2r3)
    t3 := InitNode(3, t3l1r1, t3l1r2)

    fmt.Printf("merge t1:")
    TreePrint(t1)
    fmt.Println()

    fmt.Printf("merge t2:")
    TreePrint(t2)
    fmt.Println()

    ret := MergeTrees(t1, t2)

    ok := treeEqual(t3, ret)

    fmt.Printf("merge t3:")
    TreePrint(t1)
    fmt.Println()

    fmt.Printf("t1:%+v, t2:%+v, t3:%+v\n", t1, t2, t3)
    if !ok {
        t.Errorf("fail, ret.Val want %+v, get %+v\n", t3, ret)
    } else {
        t.Logf("pass")
    }



}

C语言版:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2) {
    if (t1 && t2) {
    struct TreeNode *node = (struct TreeNode *)malloc(sizeof(struct TreeNode));
    node->val = t1->val + t2->val;
    node->left = mergeTrees(t1 ? t1->left : NULL, t2 ? t2->left : NULL);
    node->right = mergeTrees(t1 ? t1->right : NULL, t2 ? t2->right : NULL);
    return node;
    }
    return t1 ? t1 : t2;
}

C++版

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (t1 && t2) {
            TreeNode *node = new TreeNode(t1->val + t2->val);
            node->left = mergeTrees(t1 ? t1->left : nullptr, t2 ? t2->left : nullptr);
            node->right = mergeTrees(t1 ? t1->right : nullptr, t2 ? t2->right : nullptr);
            return node;
        }
        return t1 ? t1 : t2;
    }
};

Java版:

/**
 * 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;
        if (t2 == null) return t1;
        int value = (t1 == null ? 0 : t1.val) + (t2 == null ? 0 : t2.val);
        TreeNode node = new TreeNode(value);
        node.left = mergeTrees(t1 == null ? null : t1.left, t2 == null ? null : t2.left);
        node.right = mergeTrees(t1 == null ? null : t1.right, t2 == null ? null : t2.right);
        return node;
    }
}

链接:

  1. http://www.jianshu.com/p/fc41005c1e83
  2. https://leetcode.com/problems/merge-two-binary-trees/description/

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感谢作者:miltonsun

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