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| 1 | +/** |
| 2 | + * Definition for a binary tree node. |
| 3 | + * public class TreeNode { |
| 4 | + * int val; |
| 5 | + * TreeNode left; |
| 6 | + * TreeNode right; |
| 7 | + * TreeNode() {} |
| 8 | + * TreeNode(int val) { this.val = val; } |
| 9 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 10 | + * this.val = val; |
| 11 | + * this.left = left; |
| 12 | + * this.right = right; |
| 13 | + * } |
| 14 | + * } |
| 15 | + */ |
| 16 | +class Solution { |
| 17 | + public TreeNode constructFromPrePost(int[] preorder, int[] postorder) { |
| 18 | + int n = preorder.length; |
| 19 | + Map<Integer, Integer> postorderIndex = new HashMap<>(); |
| 20 | + for (int i = 0; i < n; i++) { |
| 21 | + postorderIndex.put(postorder[i], i); |
| 22 | + } |
| 23 | + return constructTree(0, n - 1, 0, preorder, postorderIndex); |
| 24 | + } |
| 25 | + |
| 26 | + private TreeNode constructTree(int preStart, int preEnd, int postStart, int[] preorder, Map<Integer, Integer> postorderIndex) { |
| 27 | + if (preStart > preEnd) { |
| 28 | + return null; |
| 29 | + } |
| 30 | + if (preStart == preEnd) { |
| 31 | + return new TreeNode(preorder[preStart]); |
| 32 | + } |
| 33 | + int leftRoot = preorder[preStart + 1]; |
| 34 | + int leftCount = postorderIndex.get(leftRoot) - postStart + 1; |
| 35 | + TreeNode root = new TreeNode(preorder[preStart]); |
| 36 | + root.left = constructTree(preStart + 1, preStart + leftCount, postStart, preorder, postorderIndex); |
| 37 | + root.right = constructTree(preStart + leftCount + 1, preEnd, postStart + leftCount, preorder, postorderIndex); |
| 38 | + return root; |
| 39 | + } |
| 40 | +} |
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