lintcode: 中序遍历和后序遍历树构造二叉树

题目

中序遍历和后序遍历树构造二叉树

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，程序来源于九章

/**
 * 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<inorder.length;i++)
            if(inorder[i] == r)
                return i;
        return -1;
    }
    public TreeNode buildTree(int[] inorder ,int istart,int iend,int[] postorder,int pstart,int pend){
        if(istart > 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;
    }
}