LintCode 中序遍历和后序遍历树构造二叉树

题目

根据中序遍历和后序遍历树构造二叉树

注意事项

你可以假设树中不存在相同数值的节点

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

代码

/**
 * 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) {
        if (inorder.length != postorder.length) {
            return null;
        }
        return myBuildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 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, root.val);
        
        
        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;
    }
    
    private int findposition(int[] inorder, int key) {
        for(int i=0;i