Java Solution with Graph

  • 0
        public boolean verifyPreorder(int[] preorder) {
          int low = Integer.MIN_VALUE;
          Stack<Integer> stack = new Stack();
          for(int p : preorder){
            if(p < low) return false;
            while(!stack.isEmpty() && p > stack.peek()){
              low = stack.pop();    
          return true;

    I cannot image the solution without drawing a real tree. In my case, it is a invalid BST, however the sequence can represented another valid BST.
    alt text

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