leedcode做题总结, 题目Construct Binary Tree from Preorder。。。105/106

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本文介绍了使用递归方法构建二叉树的两种不同场景，包括前序与中序遍历构建和后序与中序遍历构建。通过具体代码实现，详细解释了构建过程中的关键步骤。

这两题主要是用找规律的方法，找出两种序列的关系然后递归构造即可。

public class Solution {
    private TreeNode generate(int i, int j, int m, int n, int[] preorder, int[] inorder){
        if(i==j)
            return new TreeNode(preorder[i]);
        if(i>j)
            return null;
        TreeNode root = new TreeNode(preorder[i]);
        int r;
        for(r=m;r<=n;r++){
            if(inorder[r]==preorder[i]){
                root.left=generate(i+1,i+r-m,m,r-1,preorder,inorder);
                break;
            }
        }
        root.right=generate(i+r-m+1,j,r+1,n,preorder, inorder);
        return root;
    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length==0)
            return null;
        if(preorder.length==1)
            return new TreeNode(preorder[0]);

        TreeNode root = generate(0,preorder.length-1,0,inorder.length-1,preorder,inorder);
        return root;
    }
}

public class Solution {
    private TreeNode generate(int i, int j, int m, int n, int[] postorder, int[] inorder){
        if(i==j)
            return new TreeNode(postorder[i]);
        if(i>j)
            return null;
        TreeNode root = new TreeNode(postorder[i]);
        int r;
        for(r=m;r<=n;r++){
            if(inorder[r]==postorder[i]){
                root.right=generate(i+1,i+r-m,m,r-1,postorder,inorder);
                break;
            }
        }
        root.left=generate(i+r-m+1,j,r+1,n,postorder, inorder);
        return root;
    }

    public TreeNode buildTree(int[] inorder, int[] postorder) {

        if(postorder.length==0)
            return null;
        if(postorder.length==1)
            return new TreeNode(postorder[0]);
        int[] post = new int[postorder.length];
        int[] in = new int[postorder.length];
        for(int i=0;i<postorder.length;i++){
            post[i] = postorder[postorder.length-1-i];
            in[i] = inorder[postorder.length-1-i];
        }

        TreeNode root = generate(0,postorder.length-1,0,inorder.length-1,post,in);
        return root;
    }
}

Update 2015/09/01:第二题的做法有点繁琐，其实把两个数组从后往前数就相当于第一题的从右往左建立树。

/**
 * 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 TreeNode generate(int i, int j, int m, int n, int[] postorder, int[] inorder){  
        if(i==j)  
            return new TreeNode(postorder[i]);  
        if(i<j)  
            return null;  
        TreeNode root = new TreeNode(postorder[i]);  
        int r;  
        for(r=m;r>=n;r--){  
            if(inorder[r]==postorder[i]){  
                root.right=generate(i-1,i-m+r,m,r+1,postorder,inorder);  
                break;  
            }  
        }  
        root.left=generate(i-m+r-1,j,r-1,n,postorder, inorder);  
        return root;  
    }
    
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        // write your code here
        if(postorder.length==0)  
            return null;
 
        TreeNode root = generate(postorder.length-1, 0,inorder.length-1, 0,postorder,inorder);  
        return root; 
    }
}