代码随想录算法刷题训练营day19-编程知识

代码随想录算法刷题训练营day19：LeetCode(404)左叶子之和、LeetCode(112)路径总和、LeetCode(113)路径总和 II、LeetCode(105)从前序与中序遍历序列构造二叉树、LeetCode(106)从中序与后序遍历序列构造二叉树

LeetCode(404)左叶子之和
题目
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代码

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public int sumOfLeftLeaves(TreeNode root) {//int sum=0;//递归条件变化一下if(root==null){return 0;}if(root.left==null&&root.right==null){return 0;}//int sum=0;int result=sumOfLeftLeavesTest(root);return result;}public int sumOfLeftLeavesTest(TreeNode root){if(root==null){return 0;}if(root.left==null&&root.right==null){return 0;}int leftSum=sumOfLeftLeavesTest(root.left);//此时是左子树，同样也是叶子节点int rightSum=sumOfLeftLeavesTest(root.right);int leftValue=0;if(root.left!=null&&root.left.left==null&&root.left.right==null){//空指针异常问题leftValue=root.left.val;}int sum=leftSum+rightSum+leftValue;return sum;}
}

LeetCode(112)路径总和
题目
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代码

import java.util.ArrayList;
import java.util.List;/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {//采用前序遍历，把二叉树的所有路径取出来，放到集合通过遍历集合读出路径和List<List> pathDate=new ArrayList<>();List<Integer> paths=new ArrayList<>();//定义一个获取路径的函数getPaths(root,pathDate,paths);for(int i=0;i<pathDate.size();i++){List<Integer> path=pathDate.get(i);int sum=0;for(int j=0;j<path.size();j++){sum=sum+path.get(j);}if(sum==targetSum){return true;}}return false;}public void getPaths(TreeNode root,List<List> pathDate,List<Integer> paths){//设置终止条件if(root==null){return;}//前序遍历----到根节点进行判断---不然如果仅有一个节点，加入不了根结点数据paths.add(root.val);//判断到叶子节点的时候，将存储的路径数据放到集合里面if(root.left==null&&root.right==null){//将路径存储到pathDate中List<Integer> date = new ArrayList<>();date.addAll(paths);//拷贝数据----引用型数据不能直接赋值pathDate.add(date);}//遍历左子树if(root.left!=null){getPaths(root.left, pathDate, paths);//回溯paths.remove(paths.size()-1);}//遍历右子树if(root.right!=null){getPaths(root.right, pathDate, paths);//回溯paths.remove(paths.size()-1);}}
}

LeetCode(113)路径总和 II
题目
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代码

import java.util.ArrayList;
import java.util.List;/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<List<Integer>> pathSum(TreeNode root, int targetSum) {List<List> datePath=new ArrayList<>();//存放所有路径数据List<Integer> paths=new ArrayList<>();//存放遍历路径数据List<List<Integer>> resultPath=new ArrayList<>();//存放结果路径数据getTreePath(root,datePath,paths);for(int i=0;i<datePath.size();i++){List<Integer> path=new ArrayList<>();path=datePath.get(i);int sum=0;for(int j=0;j<path.size();j++){sum=sum+path.get(j);}if(sum==targetSum){resultPath.add(path);}}return resultPath;}//定义获取树路径的函数public void getTreePath(TreeNode root,List<List> datePath,List<Integer> paths){if(root==null){return;//终止条件1---防止空节点异常}//收尾路径----先通过先序遍历paths.add(root.val);//遍历根节点，保证有数据if(root.left==null&&root.right==null){List<Integer> data=new ArrayList<>();data.addAll(paths);datePath.add(data);}//遍历左子树if(root.left!=null){getTreePath(root.left, datePath, paths);//回溯paths.remove(paths.size()-1);}//遍历右子树if(root.right!=null){getTreePath(root.right, datePath, paths);//回溯paths.remove(paths.size()-1);}}
}

LeetCode(105)从前序与中序遍历序列构造二叉树
题目
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代码

import java.util.Arrays;/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public TreeNode buildTree(int[] preorder, int[] inorder) {//切割+递归//终止条件if(preorder.length==0){return null;}//取根节点int rootdate=preorder[0];TreeNode root=new TreeNode(rootdate);int index=0;//找出中序遍历数组的中间节点for (int i = 0; i < inorder.length; i++) {if(rootdate==inorder[i]){index=i;}    }//切割中序int[] leftInorder=Arrays.copyOfRange(inorder, 0, index);int[] rightInorder=Arrays.copyOfRange(inorder, index+1, inorder.length);//切割后续int dateLength=leftInorder.length;int[] leftPreorder=Arrays.copyOfRange(preorder, 1, 1+dateLength);int[] rightPreorder=Arrays.copyOfRange(preorder, 1+dateLength, preorder.length);//继续递归root.left=buildTree(leftPreorder, leftInorder);root.right=buildTree(rightPreorder, rightInorder);return root;}
}

LeetCode(106)从中序与后序遍历序列构造二叉树
题目
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代码

import java.util.Arrays;/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {//方法：进行切割中序和后序//判断树是否为空树if(inorder.length==0){return null;//终止条件}//后序遍历数组的最后一个数字是根节点//把根节点高处来，便于切割中序数组和中序数组//递归终止条件int index=0;TreeNode root=new TreeNode();int indexRoot=postorder[postorder.length-1];/* if(postorder.length>0){//设置root.val=indexRoot;//切割中序数组      } */root.val=indexRoot;for (int i = 0; i < inorder.length; i++) {if(indexRoot==inorder[i]){index=i;break;}  }  //获取中间索引位置//获取中序遍历中左子树的中序遍历int[] leftInorder=Arrays.copyOfRange(inorder, 0, index);int[] rightInorder=Arrays.copyOfRange(inorder, index+1, inorder.length);//切割后序遍历数组int leftLength=leftInorder.length;int[] leftPostorder=Arrays.copyOfRange(postorder, 0, leftLength);int[] rightPostorder=Arrays.copyOfRange(postorder, leftLength, postorder.length-1);//根据下标截取数组//递归开始，将左子树，仍进去root.left=buildTree(leftInorder, leftPostorder);root.right=buildTree(rightInorder, rightPostorder);return root;}
}