第73题 Construct Binary Tree from Preorder and Inorder Traversal(带主函数)

本文介绍如何利用预序遍历(先根节点后子节点)和中序遍历(先左子树后根节点再右子树)构建二叉树,通过HashMap辅助查找节点位置,实现递归的构建过程。

Construct Binary Tree from Preorder and Inorder Traversal

/**
 * 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 preorder: A list of integers that preorder traversal of a tree
     * @param inorder: A list of integers that inorder traversal of a tree
     * @return: Root of a tree
     */
    public void main(String[] args){
        int[] A= new int[] {2,1,3} ;
        int[] B = new int[] {1,2,3};
        TreeNode root =  buildTree(A , B) ;
        print(root) ;

    }
    public  void print(TreeNode root){
		if(root == null){
			return ;
		}
		print(root.left);
		System.out.print(root.val + " ");
		print(root.right);
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // write your code here
        HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
        if (preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0 ){
            return null ;
        }  
        for (int i = 0 ; i < inorder.length ; i++){
            map.put(inorder[i] , i);
        }
        return helper(preorder , inorder, 0 , preorder.length-1 , 0 , inorder.length-1, map );
    }
    public TreeNode helper(int[] preorder, int[] inorder, int preL, int preR, int inL, int inR, HashMap<Integer,Integer> map){
       // TreeNode root = new TreeNode();
        if(preL > preR || inL > inR){
            
            return null ;
        }
        TreeNode root = new TreeNode(preorder[preL]) ;
        int index = map.get(root.val);
        root.left = helper(preorder , inorder, preL+1 , preL+index-inL , inL, index-1 , map);
        root.right = helper(preorder, inorder,  preR-(inR-index)+1, preR , index+1 , inR, map) ;
        return root ;

    }
}
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