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前序中序/中序后续推二叉树

前序中序/中序后续推二叉树

作者: Tycc | 来源:发表于2018-03-26 20:00 被阅读0次

    前序中序推二叉树

    //我的做法,效率不是很好,花了31 ms。拷贝数组肯定很费时间。
    /**
     * 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[] preorder, int[] inorder) {
            if(preorder.length == 0 || inorder.length == 0)
                return null;
              
            int rootVal = preorder[0];
            int rootIndex = findIndex(inorder, rootVal);
            TreeNode root = new TreeNode(rootVal);
            
            int leftNum = rootIndex;
          
            if(rootIndex == 0 )
                root.left = null;
            else
                root.left = buildTree(Arrays.copyOfRange(preorder, 1, leftNum + 1), Arrays.copyOfRange(inorder, 0, rootIndex));
            
            if(rootIndex == preorder.length)
                root.right = null;
            else
                root.right = buildTree(Arrays.copyOfRange(preorder, leftNum + 1, preorder.length), Arrays.copyOfRange(inorder, rootIndex + 1, preorder.length));
            
            return root;
        }
        
        int findIndex(int[] array, int k){
            for(int i = 0; i< array.length; i++){
                if(k == array[i])
                    return i;
            }
            return -1;
        }
           
    }
    
    //detail里耗时2ms的解法
    /**
     * 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[] preorder, int[] inorder) {
            return buildTree(preorder,0,inorder,inorder.length-1,0);
            
        }
        private TreeNode buildTree(int[] preorder,int p,int[] inorder,int start,int end){
            if(start<end || p>preorder.length-1)
                return null;
            int val = preorder[p];
            int index = start;
            for(int i=start;i>=end;i--){
                if(val==inorder[i]){
                    index = i;
                    break;
                }
            }
            TreeNode node = new TreeNode(val);
            node.left = buildTree(preorder,p+1,inorder,index-1,end);
            node.right = buildTree(preorder,p+(index-end)+1,inorder,start,index+1);
            return node;
        }
    }
    

    后序中序推二叉树

    /**
     * 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) {
            return buildTree(postorder, postorder.length - 1, inorder, 0, inorder.length - 1);
        }
        
        TreeNode buildTree(int[] postorder, int p, int[] inorder, int start, int end){
            if(start > end || p < 0)
                return null;
            
            int val = postorder[p];
            int index = start;
            
            for(int i = start; i <= end; i++){
                if(val == inorder[i]){
                    index = i;
                    break;
                }
            }
            
            TreeNode node = new TreeNode(val);
            node.left = buildTree(postorder, p - (end - index) - 1, inorder, start, index - 1);
            node.right = buildTree(postorder, p-1, inorder, index + 1, end);
            return node;        
        }
        
        
    }
    

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