题目

根据中序遍历和后序遍历树构造二叉树

样例

给出树的中序遍历： [1,2,3] 和后序遍历： [1,3,2]

返回如下的树：

  2

 /  \

1    3

注意

你可以假设树中不存在相同数值的节点

解题

1.后序遍历最后一个结点就是根节点，根据这个根结点把中序遍历划分开来，同时也把后续遍历划分开来

2.递归就好了

程序感觉很简单不知道怎么写的，程序来源于

/** * 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 int findPosition(int[] arr, int start, int end, int key) {        int i;        for (i = start; i <= end; i++) {            if (arr[i] == key) {                return i;            }        }        return -1;    }    private TreeNode myBuildTree(int[] inorder, int instart, int inend,            int[] postorder, int poststart, int postend) {        if (instart > inend) {            return null;        }        TreeNode root = new TreeNode(postorder[postend]);        int position = findPosition(inorder, instart, inend, postorder[postend]);        root.left = myBuildTree(inorder, instart, position - 1,                postorder, poststart, poststart + position - instart - 1);        root.right = myBuildTree(inorder, position + 1, inend,                postorder, poststart + position - instart, postend - 1);        return root;    }    public TreeNode buildTree(int[] inorder, int[] postorder) {        if (inorder.length != postorder.length) {            return null;        }        return myBuildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);    }}

自己又实现了一遍，并加了注释

/** * 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     */    public TreeNode buildTree(int[] inorder, int[] postorder) {        // write your code here        if(inorder.length != postorder.length)            return null;        return buildTree(inorder,0,inorder.length -1 ,postorder,0,postorder.length -1 );    }    public int findroot(int[] inorder,int r){        for(int i=0;i
       
         iend)            return null;        int r = postorder[pend];        // 跟结点        TreeNode root = new TreeNode(r);        // 找到根节点        int l = findroot(inorder,r);        // 左子树 中序遍历 起始结束位置以此是：istart l-1         //后序遍历 起始位置是：pstart 结束位置:pstart(已经占据了一个位置所以要-1) + (左子树的长度) - 1        root.left = buildTree(inorder,istart,l-1,postorder,pstart,pstart+(l-1 - istart + 1) -1);        // 右子树 中序遍历 起始结束位置：l+1 iend        // 后序遍历 起始位置：pstart + (左子树的长度) ,结束位置 pend -1        root.right = buildTree(inorder,l+1,iend,postorder,pstart + (l-1-istart+1),pend -1);        return root;    }}