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.SOLUTION 1:
1. Find the root node from the preorder.(it is the first node.)
2. Try to find the position of the root in the inorder. Then we can get the number of nodes in the left tree.
3. 递归调用，构造左子树和右子树。
例子：
Pre: 4 2 1 3 6 5 7
Inorder: 1 2 3 4 5 6 7
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 public class Solution {11 public TreeNode buildTree(int[] preorder, int[] inorder) {12 // bug 3: consider when length is 0.13 if (preorder == null || inorder == null || preorder.length == 0 || preorder.length != inorder.length) {14 return null;15 }16 17 // bug 4: end index is length - 1.18 return buildTree(preorder, inorder, 0, preorder.length - 1, 0, preorder.length - 1);19 }20 21 public TreeNode buildTree(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) {22 // base case;23 if (preStart > preEnd) {24 return null;25 }26 27 int rootVal = preorder[preStart];28 TreeNode root = new TreeNode(rootVal);29 30 int pos = findTarget(inorder, rootVal, inStart, inEnd);31 32 // bug 5: left number is pos - instart can't add 133 int leftNum = pos - inStart;34 35 root.left = buildTree(preorder, inorder, preStart + 1, preStart + leftNum, inStart, pos - 1);36 root.right = buildTree(preorder, inorder, preStart + leftNum + 1, preEnd, pos + 1, inEnd);37 38 return root;39 }40 41 // bug 1: return type required.42 // bug 2: this is not a bst. can't use binary search.43 public int findTarget(int[] A, int target, int start, int end) {44 for (int i = start; i <= end; i++) {45 if (target == A[i]) {46 return i;47 }48 }49 50 return -1;51 }52 }