題目105. 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.

1,

分析,盜用一圖, 一圖勝千言

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public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || preorder.length == 0){
            return null;
        }
        
       return createTree(preorder,0,inorder,0,inorder.length-1);
    }
    
    private TreeNode createTree(int[] preorder ,int preIdx, int[] inorder, int inStartIdx, int inEndIdx){
        if(preIdx >= preorder.length){
            return null;
        }
        
        if(inStartIdx > inEndIdx){
            return null;
        }
        int rootVal = preorder[preIdx];
        TreeNode root = new TreeNode(rootVal);
        int rootIdxIn;
        for(rootIdxIn=inStartIdx; rootIdxIn<=inEndIdx; rootIdxIn++){
            if(rootVal == inorder[rootIdxIn]){
                break;
            }
        }
        root.left = createTree(preorder,preIdx+1,inorder,inStartIdx,rootIdxIn-1);
        root.right = createTree(preorder,preIdx+(rootIdxIn-inStartIdx+1),inorder,rootIdxIn+1,inEndIdx);
        return root;
    }
}