[leetcode] Construct Binary Tree from Inorder and Postorder Traversal

 

Given inorder and postorder 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-inorder-and-postorder-traversal/

 

思路：递归求解，后序遍历可以获得根节点，然后根据根节点在中序排列中的位置分出左右子树。时间复杂度O(N*N)

时间优化：可以把节点放到map中，加快在inorder中搜索的速度。 O(N)

 

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

    private TreeNode build(int[] in, int[] post, int len) {
        if (len <= 0)
            return null;
        TreeNode root = new TreeNode(post[len - 1]);
        int idx = -1;
        for (int i = 0; i < len; i++) {
            if (in[i] == post[len - 1])
                idx = i;
        }

        root.left = build(Arrays.copyOfRange(in, 0, idx), Arrays.copyOfRange(post, 0, idx), idx);
        root.right = build(Arrays.copyOfRange(in, idx + 1, len), Arrays.copyOfRange(post, idx, len - 1), len - 1 - idx);

        return root;
    }
    

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

 

时空O(N) 解法

class Solution {
    int postIdx;
    int[] postorder;
    int[] inorder;
    Map<Integer,Integer> idxMap = new HashMap<>();
    
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        this.postorder = postorder;
        this.inorder = inorder;
        postIdx = inorder.length-1;
        
        for(int i=0;i<inorder.length;i++){
            idxMap.put(inorder[i],i);
        }
        
        return helper(0, inorder.length-1);
    }
    
    public TreeNode helper(int inLeft, int inRight){
        if(inLeft>inRight){
            return null;
        }
        int rootVal = postorder[postIdx];
        TreeNode root = new TreeNode(rootVal);
        
        int idx = idxMap.get(rootVal);
        postIdx--;
        
        root.right = helper(idx+1,inRight);
        root.left = helper(inLeft,idx-1);
        return root;
    }
}

 

 

 

不创建新的数组：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null)
            return null;
        
        return build(inorder,0,inorder.length,postorder,0,postorder.length);
    }
    
    private TreeNode build(int[] in, int a, int b, int[] post, int c, int d){
        if(a>=b)
            return null;
        int rootIdx = -1;
        for(int i=a;i<b;i++){
            if(in[i]==post[d-1]){
                rootIdx = i;
            }
        }
        int leftLen = rootIdx-a;
        TreeNode root = new TreeNode(post[d-1]);
        root.left = build(in,a,rootIdx,post,c,c+leftLen);
        root.right = build(in,rootIdx+1,b,post,c+leftLen,d-1);
        
        return root;
    }
    
}

 

 利用Map O(N)算法

class Solution {
    Map<Integer,Integer> idxMap = new HashMap<>();

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null)
            return null;
        
        for(int i=0;i<inorder.length;i++){
            idxMap.put(inorder[i],i);
        }
        
        return build(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    
    private TreeNode build(int[] in, int a, int b, int[] post, int c, int d){
        if(a>b)
            return null;
        int rootIdx = idxMap.get(post[d]);

        int leftLen = rootIdx-a;
        TreeNode root = new TreeNode(post[d]);
        root.left = build(in,a,rootIdx-1,post,c,c+leftLen-1);
        root.right = build(in,rootIdx+1,b,post,c+leftLen,d-1);
        
        return root;
    }
    
}

 

 

 

 

 

 

 

第二遍记录：

　　java的写法，数组要拷贝，注意一般java边界是exclusive的

　　注意递归的边界（即数组长度为0时），开始忘记了。

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null)
            return null;
        return build(inorder,postorder);
    }
    
    private TreeNode build(int[] inorder, int[] postorder){
        if(inorder==null||inorder.length==0)
            return null;
        int rootVal = postorder[postorder.length-1];
        int idx=-1;
        for(int i=0;i<inorder.length;i++){
            if(inorder[i]==rootVal){
                idx=i;
                break;
            }
        }
        TreeNode root = new TreeNode(rootVal);
        root.left= build(Arrays.copyOfRange(inorder,0,idx),Arrays.copyOfRange(postorder,0,idx));
        root.right=build(Arrays.copyOfRange(inorder,idx+1,inorder.length),Arrays.copyOfRange(postorder,idx,postorder.length-1));
        
        return root;
    }
    
}