[leetcode] Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

 

思路：类似上题，从preorder入手找到根节点然后在中序中分辨左右子树。

　　因为java不支持返回数组后面元素的地址，所以写起来不如C/C++优雅，需要传递一个范围或者要局部复制数组。

public class Solution {

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (inorder.length == 0 || preorder.length == 0)
            return null;
        TreeNode res = build(preorder, 0, preorder.length, inorder, 0, inorder.length);
        return res;

    }

    private TreeNode build(int[] pre, int a, int b, int[] in, int c, int d) {
        if (b - a <= 0)
            return null;
        TreeNode root = new TreeNode(pre[a]);
        int idx = -1;
        for (int i = c; i < d; i++) {
            if (in[i] == pre[a])
                idx = i;
        }
        // use the len, not idx
        int len = idx - c;
        root.left = build(pre, a + 1, a + 1 + len, in, c, c + len);
        root.right = build(pre, a + 1 + len, b, in, c + 1 + len, d);
        return root;
    }

    public static void main(String[] args) {
        new Solution().buildTree(new int[] { 3, 9, 20, 15, 7 }, new int[] { 9, 3, 15, 20, 7 });
    }
}

 

 

2022 回顾，如果每次都重新create 新数组递归下去，无法简单使用map来加快inorder里index的查找，因为index是global index。所以这里采用传递数组坐标的方式递归。

 

class Solution {
    private Map<Integer,Integer> inMap = new HashMap<>();
    private int[] preorder;
    private int[] inorder;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || inorder == null){
            return null;
        }
        this.preorder = preorder;
        this.inorder = inorder;
        for(int i=0;i<inorder.length;i++){
            inMap.put(inorder[i],i);
        }
        return build(0, preorder.length, 0, inorder.length);
    }
    
    private TreeNode build(int preS, int preE, int inS, int inE){
        if(preS==preE){
            return null;
        }
        
        int rootVal = preorder[preS];
        //be careful here, the idx is the global index here
        int idx = inMap.get(rootVal);
        int leftLen = idx - inS;
        
        TreeNode root = new TreeNode(rootVal);
        root.left = build(preS+1, preS+1+leftLen, inS, idx);
        root.right = build(preS+1+leftLen, preE, idx+1, inE);
        
        return root;
    }
    
}