leedcode做题总结, 题目Construct Binary Tree from Preorder。。。105/106

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这两题主要是用找规律的方法，找出两种序列的关系然后递归构造即可。

public class Solution {    private TreeNode generate(int i, int j, int m, int n, int[] preorder, int[] inorder){        if(i==j)            return new TreeNode(preorder[i]);        if(i>j)            return null;        TreeNode root = new TreeNode(preorder[i]);        int r;        for(r=m;r<=n;r++){            if(inorder[r]==preorder[i]){                root.left=generate(i+1,i+r-m,m,r-1,preorder,inorder);                break;            }        }        root.right=generate(i+r-m+1,j,r+1,n,preorder, inorder);        return root;    }    public TreeNode buildTree(int[] preorder, int[] inorder) {        if(preorder.length==0)            return null;        if(preorder.length==1)            return new TreeNode(preorder[0]);        TreeNode root = generate(0,preorder.length-1,0,inorder.length-1,preorder,inorder);        return root;    }}

public class Solution {    private TreeNode generate(int i, int j, int m, int n, int[] postorder, int[] inorder){        if(i==j)            return new TreeNode(postorder[i]);        if(i>j)            return null;        TreeNode root = new TreeNode(postorder[i]);        int r;        for(r=m;r<=n;r++){            if(inorder[r]==postorder[i]){                root.right=generate(i+1,i+r-m,m,r-1,postorder,inorder);                break;            }        }        root.left=generate(i+r-m+1,j,r+1,n,postorder, inorder);        return root;    }    public TreeNode buildTree(int[] inorder, int[] postorder) {        if(postorder.length==0)            return null;        if(postorder.length==1)            return new TreeNode(postorder[0]);        int[] post = new int[postorder.length];        int[] in = new int[postorder.length];        for(int i=0;i<postorder.length;i++){            post[i] = postorder[postorder.length-1-i];            in[i] = inorder[postorder.length-1-i];        }        TreeNode root = generate(0,postorder.length-1,0,inorder.length-1,post,in);        return root;    }}

Update 2015/09/01:第二题的做法有点繁琐，其实把两个数组从后往前数就相当于第一题的从右往左建立树。

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */  public class Solution {    /**     *@param inorder : A list of integers that inorder traversal of a tree     *@param postorder : A list of integers that postorder traversal of a tree     *@return : Root of a tree     */     private TreeNode generate(int i, int j, int m, int n, int[] postorder, int[] inorder){          if(i==j)              return new TreeNode(postorder[i]);          if(i<j)              return null;          TreeNode root = new TreeNode(postorder[i]);          int r;          for(r=m;r>=n;r--){              if(inorder[r]==postorder[i]){                  root.right=generate(i-1,i-m+r,m,r+1,postorder,inorder);                  break;              }          }          root.left=generate(i-m+r-1,j,r-1,n,postorder, inorder);          return root;      }        public TreeNode buildTree(int[] inorder, int[] postorder) {        // write your code here        if(postorder.length==0)              return null;         TreeNode root = generate(postorder.length-1, 0,inorder.length-1, 0,postorder,inorder);          return root;     }}

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