重建二叉树

先来看两种写法：

第一种：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {                 if (pre == null || in == null ) {             return null;         }         try {             return ConstructCore(pre, 0, pre.length - 1, in, 0,in.length - 1);         } catch (InvalidPutException e) {             e.printStackTrace();             return null;         }               }         public static TreeNode ConstructCore(int[] preOrder,int startPreIndex, int endPreIndex,              int[] inOrder,int startInIndex, int endInIndex) throws InvalidPutException {            int rootValue = preOrder[startPreIndex];         System.out.println("rootValue = " + rootValue);         TreeNode root = new TreeNode(rootValue);            // 只有一个元素         if (startPreIndex == endPreIndex) {             if (startInIndex == endInIndex                     && preOrder[startPreIndex] == inOrder[startInIndex]) {                 System.out.println("only one element");                 return root;             } else {                 throw new InvalidPutException();             }         }            // 在中序遍历中找到根结点的索引         int rootInIndex = startInIndex;            while (rootInIndex <= endInIndex && inOrder[rootInIndex] != rootValue) {             ++rootInIndex;         }            if (rootInIndex == endInIndex && inOrder[rootInIndex] != rootValue) {             throw new InvalidPutException();            }            int leftLength = rootInIndex - startInIndex;            int leftPreOrderEndIndex = startPreIndex + leftLength;            if (leftLength > 0) {             // 构建左子树             root.left = ConstructCore(preOrder, startPreIndex + 1,                     leftPreOrderEndIndex, inOrder, startInIndex,                     rootInIndex - 1);         }            if (leftLength < endPreIndex - startPreIndex) {             // 右子树有元素,构建右子树             root.right = ConstructCore(preOrder, leftPreOrderEndIndex + 1,                     endPreIndex, inOrder, rootInIndex + 1, endInIndex);         }         return root;     }          static class InvalidPutException extends Exception {            private static final long serialVersionUID = 1L;        } }

第二种：

public class Solution {    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {        TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);        return root;    }    //前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}    private TreeNode reConstructBinaryTree(int [] pre,int startPre,int endPre,int [] in,int startIn,int endIn) {                 if(startPre>endPre||startIn>endIn)            return null;        TreeNode root=new TreeNode(pre[startPre]);                 for(int i=startIn;i<=endIn;i++)            if(in[i]==pre[startPre]){                root.left=reConstructBinaryTree(pre,startPre+1,startPre+i-startIn,in,startIn,i-1);                root.right=reConstructBinaryTree(pre,i-startIn+startPre+1,endPre,in,i+1,endIn);            }                         return root;    }}

显然第二种更好。显著的减少了if语句